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日清單理解

Mary-Kate Olsen
發布: 2025-01-01 14:14:10
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780 人瀏覽過

Day-List Comprehension

列表理解:

這是一種在語法上優雅的方法,可以在一行程式碼中建立或操作清單。

編寫程式列印包含字母「a」的水果(使用for循環):

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

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

print(newlist)

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

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(使用列表理解):

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

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

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['apple', 'banana', 'mango']
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寫一個程式來計算給定數字的平方根:

l = [10,20,30,40]
newlist = []

#using for 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)
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[100, 400, 900, 1600]
[100, 400, 900, 1600]
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從 2 個清單中找到相似的號碼以及從相同的 2 個清單中找到不同的號碼:

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

#for loop
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])

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30 40 
[30, 40]
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1 = [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)

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

for num in l2:
    if num not in l1:
        print(num,end=' ')
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[10, 20, 50, 60]
10 20 50 60

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以綜合方法為給定輸出編寫程式:

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)

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[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]
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為給定的輸出寫一個程式:
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))

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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)
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abc123
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為給定的輸出寫一個程式:

s = "a4k3b2"
輸出:aeknbd

s = "a4k3b2"
i = 0 
while i<len(s):
    first = s[i]
    second = int(s[i+1])
    print(first, chr(ord(first)+second),sep='',end='')
    i+=2

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aeknbd
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任務:

1) 寫一個程式來取得輸出 'abbbbklllbcc'

s = "a4k3b2"
i = 0

while i < len(s):
    first = s[i]  
    second =int(s[i + 1]) 
    print(first,chr(ord(first)+1)*second,sep='',end='')
    i+=2

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abbbbklllbcc
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2) 寫一個程式來取得輸出 'aaaaakkkkbbb'

s = "a4k3b2"
i = 0

while i < len(s):
    first = s[i]  
    second =int(s[i + 1]) 
    print(first,chr(ord(first))*second,sep='',end='')
    i+=2
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aaaaakkkkbbb

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3.使用綜合 for 和普通 for 迴圈將給定矩陣加入單一清單。

方法:1(使用for迴圈)

matrix = [[10,20,30], [40,50,60], [70,80,90]]
output=[]

for i in matrix:
    for j in i:
        output.append(j)
print(output)

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方法:2(使用綜合for循環)

matrix = [[10, 20, 30], [40, 50, 60], [70, 80, 90]]

output = [j for i in matrix for j in i]
print(output)

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[10, 20, 30, 40, 50, 60, 70, 80, 90]

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4.l = ['ABC','DEF','GHI','JKL']
取得輸出:['ABC', 'def','GHI', 'jkl']

l = ['ABC', 'DEF', 'GHI', 'JKL']

output = [] 
for i, alpha in enumerate(l):
    if i % 2 != 0:
        output.append(alpha.casefold())
    else:
        output.append(alpha)
print(output)
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['ABC', 'def', 'GHI', 'jkl']

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5.求給定矩陣的行總計:

matrix = [[10,20,30], 
          [40,50,60], 
          [70,80,90]]
for inner in matrix:
    total = 0
    for index,num in enumerate(inner):
        total+=num
    print(total,end=' ')

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60 150 240
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以上是日清單理解的詳細內容。更多資訊請關注PHP中文網其他相關文章!

來源:dev.to
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