列表理解：
这是一种在语法上优雅的方法，可以在一行代码中创建或操作列表。
编写程序打印包含字母“a”的水果（使用for循环）：
fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
newlist = []
for x in fruits:
if "a" in x:
newlist.append(x)
print(newlist)
['apple', 'banana', 'mango']
（使用列表理解）：
fruits = ["apple", "banana", "cherry", "kiwi", "mango"] newlist = [x for x in fruits if "a" in x] print(newlist)
['apple', 'banana', 'mango']
编写一个程序来计算给定数字的平方根：
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)
[100, 400, 900, 1600] [100, 400, 900, 1600]
从 2 个列表中查找相似的号码以及从相同的 2 个列表中查找不同的号码：
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])
30 40 [30, 40]
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=' ')
[10, 20, 50, 60] 10 20 50 60
以综合方法为给定输出编写程序：
l1 = [1,2,3]
l2 = [5,6,7]
输出：[(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)
[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]
为给定的输出编写一个程序：
s = "a1b2c3"
输出：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)
abc123
为给定的输出编写一个程序：
s = "a4k3b2"
输出：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
aeknbd
任务：
1) 编写一个程序来获取输出 '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
abbbbklllbcc
2) 编写一个程序来获取输出 '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
aaaaakkkkbbb
3.使用综合 for 和普通 for 循环将给定矩阵加入到单个列表中。
方法：1（使用for循环）
matrix = [[10,20,30], [40,50,60], [70,80,90]]
output=[]
for i in matrix:
for j in i:
output.append(j)
print(output)
方法：2（使用综合for循环）
matrix = [[10, 20, 30], [40, 50, 60], [70, 80, 90]] output = [j for i in matrix for j in i] print(output)
[10, 20, 30, 40, 50, 60, 70, 80, 90]
4.l = ['ABC','DEF','GHI','JKL']
获取输出：['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)
['ABC', 'def', 'GHI', 'jkl']
5.求给定矩阵的行总计：
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=' ')
60 150 240
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