最近通过http://anandology.com/python-practice-book/working-with-data.html上的问题练习Python,做到章节2.6-列表第32个问题的时候,写出了以下代码,是否有更好的代码,可以解决这个问题。原问题如下:
Problem 32: Write a function mutate to compute all words generated by
a single mutation on a given word. A mutation is defined as inserting
a character, deleting a character, replacing a character, or swapping
2 consecutive characters in a string. For simplicity consider only
letters from a to z.
我的代码如下:
#coding:utf-8
def mutate(ori_word):
result = []
characters = [chr(w) for w in xrange(97,123)]
split_word = [w for w in ori_word] #['h', 'e', 'l', 'l', 'o']
for x in xrange(0,5):
word = split_word[:]
word.pop(x)
result.append(''.join(word)) # delete any one of the character in "hello"
for c in characters:
word = split_word[:]
word[x] = c
result.append(''.join(word)) # replace any one of the character in "hello" by using a to z
for y in xrange(0,5):
word = split_word[:]
word[x] , word[y] = word[y], word[x]
result.append(''.join(word)) # swap two character in "hello"
return result
word = mutate("hello")
高洛峰2017-04-17 17:25:32
应该多利用 list comprehension
for x in xrange(0,5)
这些应该避免使用
split_word = [w for w in ori_word]
有点多此一举了 因为可以直接 iterate through 一个 str
我的答案:
def mutate(word):
return [
value for i in range(len(word)) for letter in 'abcdefghijklmnopqrstuvwxyz' for value in [
word[:i+1]+word[i+2:i:-1]+word[i+3:], # swap
word[:i]+letter+word[i:], # insert
word[:i]+letter+word[i+1:], # replace
word[:i]+word[i+1:] # remove (duplicated)
]
]