Python標準函式庫之functools/itertools/operator
- 高洛峰原創
- 2017-02-09 11:04:561787瀏覽
引言
functools, itertools, operator是Python標準庫為我們提供的簡潔函數式編程的三大模組，這可以寫出這三個簡潔標準庫為我們提供的簡潔函數式程式設計的三大模組，這可以寫出這三個簡潔標準庫為簡潔程式設計的三大模組，更合理的這三個模組，更支持標準庫為我們提供的設計式可讀的Pythonic程式碼，接下來我們透過一些example來了解三大模組的使用。
functools的使用
functools是Python中很重要的模組，它提供了一些非常有用的高階函數。高階函數是說一個可以接受函數作為參數或以函數作為返回值的函數，因為Python中函數也是對象，因此很容易支援這樣的函數式特性。
partial
>>> from functools import partial
>>> basetwo = partial(int, base=2)
>>> basetwo('10010')
18
basetwo('10010')實際上等價於呼叫int('10010', base=2)，當函數的參數個數太多的時候，可以透過使用functools.partial來創建一個新的函數來簡化邏輯從而增強程式碼的可讀性，而partial內部實際上就是透過一個簡單的閉包來實現的。
def partial(func, *args, **keywords): def newfunc(*fargs, **fkeywords): newkeywords = keywords.copy() newkeywords.update(fkeywords) return func(*args, *fargs, **newkeywords) newfunc.func = func newfunc.args = args newfunc.keywords = keywords return newfunc
partialmethod
partialmethod和partial類似，但是對於綁定一個非物件自身的方法的時候，這個時候就只能使用partialmethod了，我們透過下面這個例子來看一下兩者的差異。
from functools import partial, partialmethod
def standalone(self, a=1, b=2):
"Standalone function"
print(' called standalone with:', (self, a, b))
if self is not None:
print(' self.attr =', self.attr)
class MyClass:
"Demonstration class for functools"
def __init__(self):
self.attr = 'instance attribute'
method1 = functools.partialmethod(standalone) # 使用partialmethod
method2 = functools.partial(standalone) # 使用partial
>>> o = MyClass() >>> o.method1() called standalone with: (<__main__.MyClass object at 0x7f46d40cc550>, 1, 2) self.attr = instance attribute # 不能使用partial >>> o.method2() Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: standalone() missing 1 required positional argument: 'self'
singledispatch
雖然Python不支援同名方法允許有不同的參數類型，但是我們可以藉用singledispatch來動態指定對應的方法所接收的參數類型，而不用把參數判斷從而放到方法內部去降低程式碼的可讀性。
from functools import singledispatch
class TestClass(object):
@singledispatch
def test_method(arg, verbose=False):
if verbose:
print("Let me just say,", end=" ")
print(arg)
@test_method.register(int)
def _(arg):
print("Strength in numbers, eh?", end=" ")
print(arg)
@test_method.register(list)
def _(arg):
print("Enumerate this:")
for i, elem in enumerate(arg):
print(i, elem)
下面透過@test_method.register(int)和@test_method.register(list)指定test_method的第一個參數為int或list的時候，分別呼叫不同的方法來處理。
>>> TestClass.test_method(55555) # call @test_method.register(int)
Strength in numbers, eh? 55555
>>> TestClass.test_method([33, 22, 11]) # call @test_method.register(list)
Enumerate this:
0 33
1 22
2 11
>>> TestClass.test_method('hello world', verbose=True) # call default
Let me just say, hello world
wraps
裝飾器會遺失被裝飾函數的__name__和__doc__等屬性，可以使用@wraps來恢復。
from functools import wraps
def my_decorator(f):
@wraps(f)
def wrapper():
"""wrapper_doc"""
print('Calling decorated function')
return f()
return wrapper
@my_decorator
def example():
"""example_doc"""
print('Called example function')
>>> example.__name__ 'example' >>> example.__doc__ 'example_doc' # 尝试去掉@wraps(f)来看一下运行结果，example自身的__name__和__doc__都已经丧失了 >>> example.__name__ 'wrapper' >>> example.__doc__ 'wrapper_doc'
我們也可以使用update_wrapper來改寫
from itertools import update_wrapper def g(): ... g = update_wrapper(g, f) # equal to @wraps(f) def g(): ...
@wraps內部其實就是基於update_wrapper來實現的。
def wraps(wrapped, assigned=WRAPPER_ASSIGNMENTS, updated=WRAPPER_UPDATES): def decorator(wrapper): return update_wrapper(wrapper, wrapped=wrapped...) return decorator
lru_cache
lru_cache和singledispatch是開發中應用非常廣泛的黑魔法，接下來我們來看看lru_cache。對於重複的運算性任務，使用快取加速是非常重要的，下面我們透過一個fibonacci的例子來看一下使用lru_cache與不使用lru_cache在速度上的差異。
# clockdeco.py
import time
import functools
def clock(func):
@functools.wraps(func)
def clocked(*args, **kwargs):
t0 = time.time()
result = func(*args, **kwargs)
elapsed = time.time() - t0
name = func.__name__
arg_lst = []
if args:
arg_lst.append(', '.join(repr(arg) for arg in args))
if kwargs:
pairs = ['%s=%r' % (k, w) for k, w in sorted(kwargs.items())]
arg_lst.append(', '.join(pairs))
arg_str = ', '.join(arg_lst)
print('[%0.8fs] %s(%s) -> %r ' % (elapsed, name, arg_str, result))
return result
return clocked
不使用lru_cache
from clockdeco import clock
@clock
def fibonacci(n):
if n < 2:
return n
return fibonacci(n-2) + fibonacci(n-1)
if __name__=='__main__':
print(fibonacci(6))下面是運行結果，從運行結果可以看出fibonacci(n)會在遞歸的時候被重複計算，這是非常耗時消費資源的。
[0.00000119s] fibonacci(0) -> 0 [0.00000143s] fibonacci(1) -> 1 [0.00021172s] fibonacci(2) -> 1 [0.00000072s] fibonacci(1) -> 1 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011444s] fibonacci(2) -> 1 [0.00022793s] fibonacci(3) -> 2 [0.00055265s] fibonacci(4) -> 3 [0.00000072s] fibonacci(1) -> 1 [0.00000072s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011158s] fibonacci(2) -> 1 [0.00022268s] fibonacci(3) -> 2 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011349s] fibonacci(2) -> 1 [0.00000072s] fibonacci(1) -> 1 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00010705s] fibonacci(2) -> 1 [0.00021267s] fibonacci(3) -> 2 [0.00043225s] fibonacci(4) -> 3 [0.00076509s] fibonacci(5) -> 5 [0.00142813s] fibonacci(6) -> 8 8
使用lru_cache
import functools
from clockdeco import clock
@functools.lru_cache() # 1
@clock # 2
def fibonacci(n):
if n < 2:
return n
return fibonacci(n-2) + fibonacci(n-1)
if __name__=='__main__':
print(fibonacci(6))下面是運行結果，對於已經計算出來的結果將其放入快取。
[0.00000095s] fibonacci(0) -> 0 [0.00005770s] fibonacci(1) -> 1 [0.00015855s] fibonacci(2) -> 1 [0.00000286s] fibonacci(3) -> 2 [0.00021124s] fibonacci(4) -> 3 [0.00000191s] fibonacci(5) -> 5 [0.00024652s] fibonacci(6) -> 8 8
上面我們選用的數字還不夠大，有興趣的朋友不妨自己選擇一個較大的數字比較一下兩者在速度上的差異
total_ordering
Python2中可以透過自訂__cmp__的回傳值0/-1/1來比較物件的大小，在Python3中廢棄了__cmp__，但是我們可以透過total_ordering接著修改__lt__() , __le__() , __gt__(), __ge__(), __eq__(), __ne__( ) 等魔術方法來自訂類別的比較規則。 p.s: 若使用必須在類別內定義 __lt__() , __le__() , __gt__(), __ge__()中的一個，並且為類別新增一個__eq__() 方法。
import functools
@functools.total_ordering
class MyObject:
def __init__(self, val):
self.val = val
def __eq__(self, other):
print(' testing __eq__({}, {})'.format(
self.val, other.val))
return self.val == other.val
def __gt__(self, other):
print(' testing __gt__({}, {})'.format(
self.val, other.val))
return self.val > other.val
a = MyObject(1)
b = MyObject(2)
for expr in ['a < b', 'a <= b', 'a == b', 'a >= b', 'a > b']:
print('\n{:<6}:'.format(expr))
result = eval(expr)
print(' result of {}: {}'.format(expr, result))下面是運行結果：
a < b : testing __gt__(1, 2) testing __eq__(1, 2) result of a < b: True a <= b: testing __gt__(1, 2) result of a <= b: True a == b: testing __eq__(1, 2) result of a == b: False a >= b: testing __gt__(1, 2) testing __eq__(1, 2) result of a >= b: False a > b : testing __gt__(1, 2) result of a > b: False
itertools的使用
itertools為我們提供了非常有用的用於操作迭代對象的函數。
無限迭代器
count
count(start=0, step=1) 會回傳一個無限的整數iterator，每次增加1。可選擇提供起始編號，預設為0。
>>> from itertools import count >>> for i in zip(count(1), ['a', 'b', 'c']): ... print(i, end=' ') ... (1, 'a') (2, 'b') (3, 'c')
cycle
cycle(iterable) 會把傳入的一個序列無限重複下去，不過可以提供第二個參數就可以製定重複次數。
>>> from itertools import cycle >>> for i in zip(range(6), cycle(['a', 'b', 'c'])): ... print(i, end=' ') ... (0, 'a') (1, 'b') (2, 'c') (3, 'a') (4, 'b') (5, 'c')
repeat
repeat(object[, times]) 傳回一個元素無限重複下去的iterator，可以提供第二個參數就可以限定重複次數。
>>> from itertools import repeat
>>> for i, s in zip(count(1), repeat('over-and-over', 5)):
... print(i, s)
...
1 over-and-over
2 over-and-over
3 over-and-over
4 over-and-over
5 over-and-over
Iterators terminating on the shortest input sequence
accumulate
accumulate(iterable[, func])
>>> from itertools import accumulate >>> import operator >>> list(accumulate([1, 2, 3, 4, 5], operator.add)) [1, 3, 6, 10, 15] >>> list(accumulate([1, 2, 3, 4, 5], operator.mul)) [1, 2, 6, 24, 120]
chain🜎 chain的實作原理如下
>>> from itertools import chain >>> list(chain([1, 2, 3], ['a', 'b', 'c'])) [1, 2, 3, 'a', 'b', 'c']
chain.from_iterable
chain.from_iterable(iterable)和chain類似，但只是接收單一iterable，然後將這個iterable中的元素組合成一個iterator。
def chain(*iterables):
# chain('ABC', 'DEF') --> A B C D E F
for it in iterables:
for element in it:
yield element
實作原理也和chain類似
>>> from itertools import chain >>> list(chain.from_iterable(['ABC', 'DEF'])) ['A', 'B', 'C', 'D', 'E', 'F']
compress
compress(data, selectors)接收兩個iterable作為參數，只返回selectors中對應的元素為True的data，當data/selectors之一用盡時停止。
def from_iterable(iterables): # chain.from_iterable(['ABC', 'DEF']) --> A B C D E F for it in iterables: for element in it: yield element
zip_longest
zip_longest(*iterables, fillvalue=None)和zip類似，但是zip的缺陷是iterable中的某一個元素被歷完，整個歷遍歷都會停止，具體差異請看下面這個範例
>>> list(compress([1, 2, 3, 4, 5], [True, True, False, False, True])) [1, 2, 5]是輸出結果
zip stops early:
[(0, 0), (1, 1)]
zip_longest processes all of the values:
[(0, 0), (1, 1), (2, None)]
islice
islice(iterable, stop) or islice(iterable, start, stop[, step]) 与Python的字符串和列表切片有一些类似，只是不能对start、start和step使用负值。
>>> from itertools import islice >>> for i in islice(range(100), 0, 100, 10): ... print(i, end=' ') ... 0 10 20 30 40 50 60 70 80 90
tee
tee(iterable, n=2) 返回n个独立的iterator，n默认为2。
from itertools import islice, tee
r = islice(count(), 5)
i1, i2 = tee(r)
print('i1:', list(i1))
print('i2:', list(i2))
for i in r:
print(i, end=' ')
if i > 1:
break
下面是输出结果，注意tee(r)后，r作为iterator已经失效，所以for循环没有输出值。
i1: [0, 1, 2, 3, 4] i2: [0, 1, 2, 3, 4]
starmap
starmap(func, iterable)假设iterable将返回一个元组流，并使用这些元组作为参数调用func：
>>> from itertools import starmap
>>> import os
>>> iterator = starmap(os.path.join,
... [('/bin', 'python'), ('/usr', 'bin', 'java'),
... ('/usr', 'bin', 'perl'), ('/usr', 'bin', 'ruby')])
>>> list(iterator)
['/bin/python', '/usr/bin/java', '/usr/bin/perl', '/usr/bin/ruby']
filterfalse
filterfalse(predicate, iterable) 与filter()相反，返回所有predicate返回False的元素。
itertools.filterfalse(is_even, itertools.count()) => 1, 3, 5, 7, 9, 11, 13, 15, ...
takewhile
takewhile(predicate, iterable) 只要predicate返回True，不停地返回iterable中的元素。一旦predicate返回False，iteration将结束。
def less_than_10(x): return x < 10 itertools.takewhile(less_than_10, itertools.count()) => 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 itertools.takewhile(is_even, itertools.count()) => 0
dropwhile
dropwhile(predicate, iterable) 在predicate返回True时舍弃元素，然后返回其余迭代结果。
itertools.dropwhile(less_than_10, itertools.count()) => 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, ... itertools.dropwhile(is_even, itertools.count()) => 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
groupby
groupby(iterable, key=None) 把iterator中相邻的重复元素挑出来放在一起。p.s: The input sequence needs to be sorted on the key value in order for the groupings to work out as expected.
[k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
[list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
>>> import itertools
>>> for key, group in itertools.groupby('AAAABBBCCDAABBB'):
... print(key, list(group))
...
A ['A', 'A', 'A', 'A']
B ['B', 'B', 'B']
C ['C', 'C']
D ['D']
A ['A', 'A']
B ['B', 'B', 'B']
city_list = [('Decatur', 'AL'), ('Huntsville', 'AL'), ('Selma', 'AL'),
('Anchorage', 'AK'), ('Nome', 'AK'),
('Flagstaff', 'AZ'), ('Phoenix', 'AZ'), ('Tucson', 'AZ'),
...
]
def get_state(city_state):
return city_state[1]
itertools.groupby(city_list, get_state) =>
('AL', iterator-1),
('AK', iterator-2),
('AZ', iterator-3), ...
iterator-1 => ('Decatur', 'AL'), ('Huntsville', 'AL'), ('Selma', 'AL')
iterator-2 => ('Anchorage', 'AK'), ('Nome', 'AK')
iterator-3 => ('Flagstaff', 'AZ'), ('Phoenix', 'AZ'), ('Tucson', 'AZ')
Combinatoric generators
product
product(*iterables, repeat=1)
product(A, B) returns the same as ((x,y) for x in A for y in B)
product(A, repeat=4) means the same as product(A, A, A, A)
from itertools import product
def show(iterable):
for i, item in enumerate(iterable, 1):
print(item, end=' ')
if (i % 3) == 0:
print()
print()
print('Repeat 2:\n')
show(product(range(3), repeat=2))
print('Repeat 3:\n')
show(product(range(3), repeat=3))
Repeat 2: (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) (2, 0) (2, 1) (2, 2) Repeat 3: (0, 0, 0) (0, 0, 1) (0, 0, 2) (0, 1, 0) (0, 1, 1) (0, 1, 2) (0, 2, 0) (0, 2, 1) (0, 2, 2) (1, 0, 0) (1, 0, 1) (1, 0, 2) (1, 1, 0) (1, 1, 1) (1, 1, 2) (1, 2, 0) (1, 2, 1) (1, 2, 2) (2, 0, 0) (2, 0, 1) (2, 0, 2) (2, 1, 0) (2, 1, 1) (2, 1, 2) (2, 2, 0) (2, 2, 1) (2, 2, 2)
permutations
permutations(iterable, r=None)返回长度为r的所有可能的组合。
from itertools import permutations
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('All permutations:\n')
show(permutations('abcd'))
print('\nPairs:\n')
show(permutations('abcd', r=2))
下面是输出结果
All permutations: abcd abdc acbd acdb adbc adcb bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc dacb dbac dbca dcab dcba Pairs: ab ac ad ba bc bd ca cb cd da db dc
combinations
combinations(iterable, r) 返回一个iterator，提供iterable中所有元素可能组合的r元组。每个元组中的元素保持与iterable返回的顺序相同。下面的实例中，不同于上面的permutations，a总是在bcd之前，b总是在cd之前，c总是在d之前。
from itertools import combinations
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('Unique pairs:\n')
show(combinations('abcd', r=2))
下面是输出结果
Unique pairs: ab ac ad bc bd cd
combinations_with_replacement
combinations_with_replacement(iterable, r)函数放宽了一个不同的约束：元素可以在单个元组中重复，即可以出现aa/bb/cc/dd等组合。
from itertools import combinations_with_replacement
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('Unique pairs:\n')
show(combinations_with_replacement('abcd', r=2))
下面是输出结果
aa ab ac ad bb bc bd cc cd dd
operator的使用
attrgetter
operator.attrgetter(attr)和operator.attrgetter(*attrs)
After f = attrgetter('name'), the call f(b) returns b.name.
After f = attrgetter('name', 'date'), the call f(b) returns (b.name, b.date).
After f = attrgetter('name.first', 'name.last'), the call f(b) returns (b.name.first, b.name.last).
我们通过下面这个例子来了解一下itergetter的用法。
>>> class Student:
... def __init__(self, name, grade, age):
... self.name = name
... self.grade = grade
... self.age = age
... def __repr__(self):
... return repr((self.name, self.grade, self.age))
>>> student_objects = [
... Student('john', 'A', 15),
... Student('jane', 'B', 12),
... Student('dave', 'B', 10),
... ]
>>> sorted(student_objects, key=lambda student: student.age) # 传统的lambda做法
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
>>> from operator import itemgetter, attrgetter
>>> sorted(student_objects, key=attrgetter('age'))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
# 但是如果像下面这样接受双重比较，Python脆弱的lambda就不适用了
>>> sorted(student_objects, key=attrgetter('grade', 'age'))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]
attrgetter的实现原理：
def attrgetter(*items):
if any(not isinstance(item, str) for item in items):
raise TypeError('attribute name must be a string')
if len(items) == 1:
attr = items[0]
def g(obj):
return resolve_attr(obj, attr)
else:
def g(obj):
return tuple(resolve_attr(obj, attr) for attr in items)
return g
def resolve_attr(obj, attr):
for name in attr.split("."):
obj = getattr(obj, name)
return obj
itemgetter
operator.itemgetter(item)和operator.itemgetter(*items)
After f = itemgetter(2), the call f(r) returns r[2].
After g = itemgetter(2, 5, 3), the call g(r) returns (r[2], r[5], r[3]).
我们通过下面这个例子来了解一下itergetter的用法
>>> student_tuples = [
... ('john', 'A', 15),
... ('jane', 'B', 12),
... ('dave', 'B', 10),
... ]
>>> sorted(student_tuples, key=lambda student: student[2]) # 传统的lambda做法
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
>>> from operator import attrgetter
>>> sorted(student_tuples, key=itemgetter(2))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
# 但是如果像下面这样接受双重比较，Python脆弱的lambda就不适用了
>>> sorted(student_tuples, key=itemgetter(1,2))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]
itemgetter的实现原理
def itemgetter(*items): if len(items) == 1: item = items[0] def g(obj): return obj[item] else: def g(obj): return tuple(obj[item] for item in items) return g
methodcaller
operator.methodcaller(name[, args...])
After f = methodcaller('name'), the call f(b) returns b.name().
After f = methodcaller('name', 'foo', bar=1), the call f(b) returns b.name('foo', bar=1).
methodcaller的实现原理
def methodcaller(name, *args, **kwargs): def caller(obj): return getattr(obj, name)(*args, **kwargs) return caller
References
DOCUMENTATION-FUNCTOOLS
DOCUMENTATION-ITERTOOLS
DOCUMENTATION-OPERATOR
HWOTO-FUNCTIONAL
HWOTO-SORTING
PYMOTW
FLENT-PYTHON
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引言
functools, itertools, operator是Python标准库为我们提供的支持函数式编程的三大模块，合理的使用这三个模块，我们可以写出更加简洁可读的Pythonic代码，接下来我们通过一些example来了解三大模块的使用。
functools的使用
functools是Python中很重要的模块，它提供了一些非常有用的高阶函数。高阶函数就是说一个可以接受函数作为参数或者以函数作为返回值的函数，因为Python中函数也是对象，因此很容易支持这样的函数式特性。
partial
>>> from functools import partial
>>> basetwo = partial(int, base=2)
>>> basetwo('10010')
18
basetwo('10010')實際上等價於調用int('10010', base=2)，當函數的參數數量太多的時候，可以透過使用functools.partial創建一個新的函數來簡化邏輯從而增強程式碼的可讀性，而partial內部實際上就是透過一個簡單的閉包來實現的。
def partial(func, *args, **keywords): def newfunc(*fargs, **fkeywords): newkeywords = keywords.copy() newkeywords.update(fkeywords) return func(*args, *fargs, **newkeywords) newfunc.func = func newfunc.args = args newfunc.keywords = keywords return newfunc
partialmethod
partialmethod和partial類似，但是對於綁定一個非物件自身的方法的時候，這個時候就只能使用partialmethod了，我們透過下面這個例子來看一下兩者的差異。
from functools import partial, partialmethod
def standalone(self, a=1, b=2):
"Standalone function"
print(' called standalone with:', (self, a, b))
if self is not None:
print(' self.attr =', self.attr)
class MyClass:
"Demonstration class for functools"
def __init__(self):
self.attr = 'instance attribute'
method1 = functools.partialmethod(standalone) # 使用partialmethod
method2 = functools.partial(standalone) # 使用partial
>>> o = MyClass() >>> o.method1() called standalone with: (<__main__.MyClass object at 0x7f46d40cc550>, 1, 2) self.attr = instance attribute # 不能使用partial >>> o.method2() Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: standalone() missing 1 required positional argument: 'self'
singledispatch
雖然Python不支援同名方法允許有不同的參數類型，但是我們可以藉用singledispatch來動態指定對應的方法所接收的參數類型，而不用把參數判斷從而放到方法內部去降低程式碼的可讀性。
from functools import singledispatch
class TestClass(object):
@singledispatch
def test_method(arg, verbose=False):
if verbose:
print("Let me just say,", end=" ")
print(arg)
@test_method.register(int)
def _(arg):
print("Strength in numbers, eh?", end=" ")
print(arg)
@test_method.register(list)
def _(arg):
print("Enumerate this:")
for i, elem in enumerate(arg):
print(i, elem)
下面透過@test_method.register(int)和@test_method.register(list)指定test_method的第一個參數為int或list的時候，分別呼叫不同的方法來處理。
>>> TestClass.test_method(55555) # call @test_method.register(int)
Strength in numbers, eh? 55555
>>> TestClass.test_method([33, 22, 11]) # call @test_method.register(list)
Enumerate this:
0 33
1 22
2 11
>>> TestClass.test_method('hello world', verbose=True) # call default
Let me just say, hello world
wraps
裝飾器會遺失被裝飾函數的__name__和__doc__等屬性，可以使用@wraps來恢復。
from functools import wraps
def my_decorator(f):
@wraps(f)
def wrapper():
"""wrapper_doc"""
print('Calling decorated function')
return f()
return wrapper
@my_decorator
def example():
"""example_doc"""
print('Called example function')
>>> example.__name__ 'example' >>> example.__doc__ 'example_doc' # 尝试去掉@wraps(f)来看一下运行结果，example自身的__name__和__doc__都已经丧失了 >>> example.__name__ 'wrapper' >>> example.__doc__ 'wrapper_doc'
我們也可以使用update_wrapper來改寫
from itertools import update_wrapper def g(): ... g = update_wrapper(g, f) # equal to @wraps(f) def g(): ...
@wraps內部其實就是基於update_wrapper來實現的。
def wraps(wrapped, assigned=WRAPPER_ASSIGNMENTS, updated=WRAPPER_UPDATES): def decorator(wrapper): return update_wrapper(wrapper, wrapped=wrapped...) return decorator
lru_cache
lru_cache和singledispatch是開發中應用非常廣泛的黑魔法，接下來我們來看看lru_cache。對於重複的運算性任務，使用快取加速是非常重要的，下面我們透過一個fibonacci的例子來看一下使用lru_cache與不使用lru_cache在速度上的差異。
# clockdeco.py
import time
import functools
def clock(func):
@functools.wraps(func)
def clocked(*args, **kwargs):
t0 = time.time()
result = func(*args, **kwargs)
elapsed = time.time() - t0
name = func.__name__
arg_lst = []
if args:
arg_lst.append(', '.join(repr(arg) for arg in args))
if kwargs:
pairs = ['%s=%r' % (k, w) for k, w in sorted(kwargs.items())]
arg_lst.append(', '.join(pairs))
arg_str = ', '.join(arg_lst)
print('[%0.8fs] %s(%s) -> %r ' % (elapsed, name, arg_str, result))
return result
return clocked
不使用lru_cache
from clockdeco import clock
@clock
def fibonacci(n):
if n < 2:
return n
return fibonacci(n-2) + fibonacci(n-1)
if __name__=='__main__':
print(fibonacci(6))下面是運行結果，從運行結果可以看出fibonacci(n)會在遞歸的時候被重複計算，這是非常耗時消費資源的。
[0.00000119s] fibonacci(0) -> 0 [0.00000143s] fibonacci(1) -> 1 [0.00021172s] fibonacci(2) -> 1 [0.00000072s] fibonacci(1) -> 1 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011444s] fibonacci(2) -> 1 [0.00022793s] fibonacci(3) -> 2 [0.00055265s] fibonacci(4) -> 3 [0.00000072s] fibonacci(1) -> 1 [0.00000072s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011158s] fibonacci(2) -> 1 [0.00022268s] fibonacci(3) -> 2 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00011349s] fibonacci(2) -> 1 [0.00000072s] fibonacci(1) -> 1 [0.00000095s] fibonacci(0) -> 0 [0.00000095s] fibonacci(1) -> 1 [0.00010705s] fibonacci(2) -> 1 [0.00021267s] fibonacci(3) -> 2 [0.00043225s] fibonacci(4) -> 3 [0.00076509s] fibonacci(5) -> 5 [0.00142813s] fibonacci(6) -> 8 8
使用lru_cache
import functools
from clockdeco import clock
@functools.lru_cache() # 1
@clock # 2
def fibonacci(n):
if n < 2:
return n
return fibonacci(n-2) + fibonacci(n-1)
if __name__=='__main__':
print(fibonacci(6))下面是運行結果，對於已經計算出來的結果將其放入快取。
[0.00000095s] fibonacci(0) -> 0 [0.00005770s] fibonacci(1) -> 1 [0.00015855s] fibonacci(2) -> 1 [0.00000286s] fibonacci(3) -> 2 [0.00021124s] fibonacci(4) -> 3 [0.00000191s] fibonacci(5) -> 5 [0.00024652s] fibonacci(6) -> 8 8
上面我們選用的數字還不夠大，有興趣的朋友不妨自己選擇一個較大的數字比較一下兩者在速度上的差異
total_ordering
Python2中可以透過自訂__cmp__的回傳值0/-1/1來比較物件的大小，在Python3中廢棄了__cmp__，但是我們可以透過total_ordering接著修改__lt__() , __le__() , __gt__(), __ge__(), __eq__(), __ne__( ) 等魔術方法來自訂類別的比較規則。 p.s: 若使用必須在類別內定義 __lt__() , __le__() , __gt__(), __ge__()中的一個，並且為類別新增一個__eq__() 方法。
import functools
@functools.total_ordering
class MyObject:
def __init__(self, val):
self.val = val
def __eq__(self, other):
print(' testing __eq__({}, {})'.format(
self.val, other.val))
return self.val == other.val
def __gt__(self, other):
print(' testing __gt__({}, {})'.format(
self.val, other.val))
return self.val > other.val
a = MyObject(1)
b = MyObject(2)
for expr in ['a < b', 'a <= b', 'a == b', 'a >= b', 'a > b']:
print('\n{:<6}:'.format(expr))
result = eval(expr)
print(' result of {}: {}'.format(expr, result))下面是運行結果：
a < b : testing __gt__(1, 2) testing __eq__(1, 2) result of a < b: True a <= b: testing __gt__(1, 2) result of a <= b: True a == b: testing __eq__(1, 2) result of a == b: False a >= b: testing __gt__(1, 2) testing __eq__(1, 2) result of a >= b: False a > b : testing __gt__(1, 2) result of a > b: False
itertools的使用
itertools為我們提供了非常有用的用於操作迭代對象的函數。
無限迭代器
count
count(start=0, step=1) 會回傳一個無限的整數iterator，每次增加1。可選擇提供起始編號，預設為0。
>>> from itertools import count >>> for i in zip(count(1), ['a', 'b', 'c']): ... print(i, end=' ') ... (1, 'a') (2, 'b') (3, 'c')
cycle
cycle(iterable) 會把傳入的一個序列無限重複下去，不過可以提供第二個參數就可以製定重複次數。
>>> from itertools import cycle >>> for i in zip(range(6), cycle(['a', 'b', 'c'])): ... print(i, end=' ') ... (0, 'a') (1, 'b') (2, 'c') (3, 'a') (4, 'b') (5, 'c')
repeat
repeat(object[, times]) 傳回一個元素無限重複下去的iterator，可以提供第二個參數就可以限定重複次數。
>>> from itertools import repeat
>>> for i, s in zip(count(1), repeat('over-and-over', 5)):
... print(i, s)
...
1 over-and-over
2 over-and-over
3 over-and-over
4 over-and-over
5 over-and-over
Iterators terminating on the shortest input sequence
accumulate
accumulate(iterable[, func])
>>> from itertools import accumulate >>> import operator >>> list(accumulate([1, 2, 3, 4, 5], operator.add)) [1, 3, 6, 10, 15] >>> list(accumulate([1, 2, 3, 4, 5], operator.mul)) [1, 2, 6, 24, 120]
chain🜎 chain的實作原理如下
>>> from itertools import chain >>> list(chain([1, 2, 3], ['a', 'b', 'c'])) [1, 2, 3, 'a', 'b', 'c']
chain.from_iterable
chain.from_iterable(iterable)和chain類似，但只是接收單一iterable，然後將這個iterable中的元素組合成一個iterator。
def chain(*iterables):
# chain('ABC', 'DEF') --> A B C D E F
for it in iterables:
for element in it:
yield element
實作原理也和chain類似
>>> from itertools import chain >>> list(chain.from_iterable(['ABC', 'DEF'])) ['A', 'B', 'C', 'D', 'E', 'F']
compress
compress(data, selectors)接收兩個iterable作為參數，只返回selectors中對應的元素為True的data，當data/selectors之一用盡時停止。
def from_iterable(iterables): # chain.from_iterable(['ABC', 'DEF']) --> A B C D E F for it in iterables: for element in it: yield element
zip_longest
zip_longest(*iterables, fillvalue=None)和zip類似，但是zip的缺陷是iterable中的某一個元素被歷完，整個歷遍歷都會停止，具體差異請看下面這個範例
>>> list(compress([1, 2, 3, 4, 5], [True, True, False, False, True])) [1, 2, 5]是輸出結果
from itertools import zip_longest
r1 = range(3)
r2 = range(2)
print('zip stops early:')
print(list(zip(r1, r2)))
r1 = range(3)
r2 = range(2)
print('\nzip_longest processes all of the values:')
print(list(zip_longest(r1, r2)))
isliceislice(iterable, stop) or islice(iterable, start, stop[, step]) 與Python的字串和列表切片有一些類似，只是不能對start、start和step使用負值。
zip stops early: [(0, 0), (1, 1)] zip_longest processes all of the values: [(0, 0), (1, 1), (2, None)]teetee(iterable, n=2) 回傳n個獨立的iterator，n預設為2。
>>> from itertools import islice >>> for i in islice(range(100), 0, 100, 10): ... print(i, end=' ') ... 0 10 20 30 40 50 60 70 80 90下面是輸出結果，注意tee(r)後，r作為iterator已經失效，所以for迴圈沒有輸出值。
from itertools import islice, tee
r = islice(count(), 5)
i1, i2 = tee(r)
print('i1:', list(i1))
print('i2:', list(i2))
for i in r:
print(i, end=' ')
if i > 1:
break
starmapstarmap(func, iterable)假設iterable會傳回一個元組流，並使用這些元組作為參數呼叫func：
i1: [0, 1, 2, 3, 4] i2: [0, 1, 2, 3, 4]filterfalsefilterfalse(preateterpter所有predicate傳回False的元素。
itertools.filterfalse(is_even, itertools.count()) => 1, 3, 5, 7, 9, 11, 13, 15, ...
takewhile
takewhile(predicate, iterable) 只要predicate返回True，不停地返回iterable中的元素。一旦predicate返回False，iteration将结束。
def less_than_10(x): return x < 10 itertools.takewhile(less_than_10, itertools.count()) => 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 itertools.takewhile(is_even, itertools.count()) => 0
dropwhile
dropwhile(predicate, iterable) 在predicate返回True时舍弃元素，然后返回其余迭代结果。
itertools.dropwhile(less_than_10, itertools.count()) => 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, ... itertools.dropwhile(is_even, itertools.count()) => 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
groupby
groupby(iterable, key=None) 把iterator中相邻的重复元素挑出来放在一起。p.s: The input sequence needs to be sorted on the key value in order for the groupings to work out as expected.
[k for k, g in groupby('AAAABBBCCDAABBB')] --> A B C D A B
[list(g) for k, g in groupby('AAAABBBCCD')] --> AAAA BBB CC D
>>> import itertools
>>> for key, group in itertools.groupby('AAAABBBCCDAABBB'):
... print(key, list(group))
...
A ['A', 'A', 'A', 'A']
B ['B', 'B', 'B']
C ['C', 'C']
D ['D']
A ['A', 'A']
B ['B', 'B', 'B']
city_list = [('Decatur', 'AL'), ('Huntsville', 'AL'), ('Selma', 'AL'),
('Anchorage', 'AK'), ('Nome', 'AK'),
('Flagstaff', 'AZ'), ('Phoenix', 'AZ'), ('Tucson', 'AZ'),
...
]
def get_state(city_state):
return city_state[1]
itertools.groupby(city_list, get_state) =>
('AL', iterator-1),
('AK', iterator-2),
('AZ', iterator-3), ...
iterator-1 => ('Decatur', 'AL'), ('Huntsville', 'AL'), ('Selma', 'AL')
iterator-2 => ('Anchorage', 'AK'), ('Nome', 'AK')
iterator-3 => ('Flagstaff', 'AZ'), ('Phoenix', 'AZ'), ('Tucson', 'AZ')
Combinatoric generators
product
product(*iterables, repeat=1)
product(A, B) returns the same as ((x,y) for x in A for y in B)
product(A, repeat=4) means the same as product(A, A, A, A)
from itertools import product
def show(iterable):
for i, item in enumerate(iterable, 1):
print(item, end=' ')
if (i % 3) == 0:
print()
print()
print('Repeat 2:\n')
show(product(range(3), repeat=2))
print('Repeat 3:\n')
show(product(range(3), repeat=3))
Repeat 2: (0, 0) (0, 1) (0, 2) (1, 0) (1, 1) (1, 2) (2, 0) (2, 1) (2, 2) Repeat 3: (0, 0, 0) (0, 0, 1) (0, 0, 2) (0, 1, 0) (0, 1, 1) (0, 1, 2) (0, 2, 0) (0, 2, 1) (0, 2, 2) (1, 0, 0) (1, 0, 1) (1, 0, 2) (1, 1, 0) (1, 1, 1) (1, 1, 2) (1, 2, 0) (1, 2, 1) (1, 2, 2) (2, 0, 0) (2, 0, 1) (2, 0, 2) (2, 1, 0) (2, 1, 1) (2, 1, 2) (2, 2, 0) (2, 2, 1) (2, 2, 2)
permutations
permutations(iterable, r=None)返回长度为r的所有可能的组合。
from itertools import permutations
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('All permutations:\n')
show(permutations('abcd'))
print('\nPairs:\n')
show(permutations('abcd', r=2))
下面是输出结果
All permutations: abcd abdc acbd acdb adbc adcb bacd badc bcad bcda bdac bdca cabd cadb cbad cbda cdab cdba dabc dacb dbac dbca dcab dcba Pairs: ab ac ad ba bc bd ca cb cd da db dc
combinations
combinations(iterable, r) 返回一个iterator，提供iterable中所有元素可能组合的r元组。每个元组中的元素保持与iterable返回的顺序相同。下面的实例中，不同于上面的permutations，a总是在bcd之前，b总是在cd之前，c总是在d之前。
from itertools import combinations
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('Unique pairs:\n')
show(combinations('abcd', r=2))
下面是输出结果
Unique pairs: ab ac ad bc bd cd
combinations_with_replacement
combinations_with_replacement(iterable, r)函数放宽了一个不同的约束：元素可以在单个元组中重复，即可以出现aa/bb/cc/dd等组合。
from itertools import combinations_with_replacement
def show(iterable):
first = None
for i, item in enumerate(iterable, 1):
if first != item[0]:
if first is not None:
print()
first = item[0]
print(''.join(item), end=' ')
print()
print('Unique pairs:\n')
show(combinations_with_replacement('abcd', r=2))
下面是输出结果
aa ab ac ad bb bc bd cc cd dd
operator的使用
attrgetter
operator.attrgetter(attr)和operator.attrgetter(*attrs)
After f = attrgetter('name'), the call f(b) returns b.name.
After f = attrgetter('name', 'date'), the call f(b) returns (b.name, b.date).
After f = attrgetter('name.first', 'name.last'), the call f(b) returns (b.name.first, b.name.last).
我们通过下面这个例子来了解一下itergetter的用法。
>>> class Student:
... def __init__(self, name, grade, age):
... self.name = name
... self.grade = grade
... self.age = age
... def __repr__(self):
... return repr((self.name, self.grade, self.age))
>>> student_objects = [
... Student('john', 'A', 15),
... Student('jane', 'B', 12),
... Student('dave', 'B', 10),
... ]
>>> sorted(student_objects, key=lambda student: student.age) # 传统的lambda做法
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
>>> from operator import itemgetter, attrgetter
>>> sorted(student_objects, key=attrgetter('age'))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
# 但是如果像下面这样接受双重比较，Python脆弱的lambda就不适用了
>>> sorted(student_objects, key=attrgetter('grade', 'age'))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]
attrgetter的实现原理：
def attrgetter(*items):
if any(not isinstance(item, str) for item in items):
raise TypeError('attribute name must be a string')
if len(items) == 1:
attr = items[0]
def g(obj):
return resolve_attr(obj, attr)
else:
def g(obj):
return tuple(resolve_attr(obj, attr) for attr in items)
return g
def resolve_attr(obj, attr):
for name in attr.split("."):
obj = getattr(obj, name)
return obj
itemgetter
operator.itemgetter(item)和operator.itemgetter(*items)
After f = itemgetter(2), the call f(r) returns r[2].
After g = itemgetter(2, 5, 3), the call g(r) returns (r[2], r[5], r[3]).
我们通过下面这个例子来了解一下itergetter的用法
>>> student_tuples = [
... ('john', 'A', 15),
... ('jane', 'B', 12),
... ('dave', 'B', 10),
... ]
>>> sorted(student_tuples, key=lambda student: student[2]) # 传统的lambda做法
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
>>> from operator import attrgetter
>>> sorted(student_tuples, key=itemgetter(2))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
# 但是如果像下面这样接受双重比较，Python脆弱的lambda就不适用了
>>> sorted(student_tuples, key=itemgetter(1,2))
[('john', 'A', 15), ('dave', 'B', 10), ('jane', 'B', 12)]
itemgetter的实现原理
def itemgetter(*items): if len(items) == 1: item = items[0] def g(obj): return obj[item] else: def g(obj): return tuple(obj[item] for item in items) return g
methodcaller
operator.methodcaller(name[, args...])
After f = methodcaller('name'), the call f(b) returns b.name().
After f = methodcaller('name', 'foo', bar=1), the call f(b) returns b.name('foo', bar=1).
methodcaller的实现原理
def methodcaller(name, *args, **kwargs): def caller(obj): return getattr(obj, name)(*args, **kwargs) return caller
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