Python Day-List 理解-練習

列表理解

當您想要基於現有清單的值建立新清單時，清單理解提供了更短的語法。 （參考-https://www.w3schools.com/python/python_lists_compressive.asp）

範例：1
方法：1

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
newlist = []

for x in fruits:
  if "a" in x:
    newlist.append(x)

print(newlist)

方法：2（綜合）

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]

newlist = [x for x in fruits if "a" in x]
print(newlist)

輸出：

['apple', 'banana', 'mango']

範例：2

l = [10,20,30,40]
newlist = []
#using normal loop
for num in l:
    newlist.append(num**2)
print(newlist)

#using loop in comprehensive way
newlist = [num**2 for num in l]
print(newlist)

輸出：

[100, 400, 900, 1600]
[100, 400, 900, 1600]

練習：
1.從2個清單中找出相似的數字以及從相同的2個清單中找出不同的數字。
l1 = [10,20,30,40]
l2 = [30,40,50,60]
得到這個輸出：
a) 30,40

#30,40
l1 = [10,20,30,40]
l2 = [30,40,50,60]
#normal method

for num in l1:
    for no in l2:
        if num== no:
            print(num,end=' ')
#comprehensive

print([num for num in l1 for no in l2 if num==no])

輸出：

[30, 40]

b) 10,20,50,60

l1 = [10,20,30,40]
l2 = [30,40,50,60]
#comprehensive
output = [num for num in l1 if num not in l2]

output = output + [num for num in l2 if num not in l1]
print(output)

#normal method
for num in l1:
    if num not in l2:
        print(num,end=' ')

for num in l2:
    if num not in l1:
        print(num,end=' ')

輸出：

[10, 20, 50, 60]
10 20 50 60 

2。以綜合方法找出給定輸出的程式
l1 = [1,2,3]
l2 = [5,6,7]
輸出：[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6) , (3, 7)]

l1 = [1,2,3]
l2 = [5,6,7]

l = [(i,j) for i in l1 for j in l2 if i!=j]
print(l)

輸出：

[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]

3。尋找給定輸出的程式：
s = "a1b2c3"
輸出：abc123

方法：1

s = "a1b2c3"

alpha_list = []
num_list = []

for letter in s:
    if letter.isalpha():
        alpha_list.append(letter)
    else:
        num_list.append(letter)

print("".join(alpha_list+num_list))

方法：2

s = "a1b2c3"
letter=''.join([i for i in s if i.isalpha()])
no=''.join([i for i in s if i.isdigit()])

print(letter+no)

輸出：

abc123

4。尋找給定輸出的程式：

s = "a4k3b2"
輸出：aeknbd

s = "a4k3b2"
i = 0 
while i<len first="s[i]" second="int(s[i+1])" print chr i>



<p>輸出：</p>

<p>aeknbd</p>

<p><strong>說明：</strong></p>

<p>-->使用ord(first)獲得first的ASCII值，然後將其新增至其中以找到新字元。 <br>
-->ord() 用來找出 ASCII 值。 <br>
-->chr() 轉換 ASCII 值-->字元。 </p>


          

            
        </len>

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