python 当外循环索引等于内循环长度时嵌套for循环你重新开始

Question

我有

A = ['A','B','C','D','E','F','G','H']
B= ['a','b']

想遍历列表得到

C = ['Aa','Bb','Ca','Db','Ea','Fb','Ga','Hb']

我怎样才能在 python

Answer 1

您可以使用列表理解：

a = ['A','B','C','D','E','F','G','H']
b = ['a','b']

c = [a[i] + b[i%2] for i in range(len(a))]

# ['Aa', 'Bb', 'Ca', 'Db', 'Ea', 'Fb', 'Ga', 'Hb']

Answer 2

您可以使用 A 列表中的索引映射到 B 列表中的元素数并进行取模运算 (%) 来完成此操作：

A = [1,2,3,4,5,6,7,8]
B= ['a','b']
C = []

for i in range(len(A)):
    C.append(f"{A[i]}{B[i%2]}")

Answer 3

您可以使用 zip() 和 cycle():

from itertools import cycle

A = ['A','B','C','D','E','F','G','H']
B = ['a', 'b']

C = [a + b for a,b in zip(A, cycle(B))]
print(C)

按要求输出。

Answer 4

A = [1, 2, 3, 4, 5, 6, 7, 8]
B = ["a", "b", "c", "d", "f", "g", "h", "q"]
C = []
i = 0
while i < len(A):
    for j in B:
        C.append("{}{}".format(A[i], j))   
    i = i + 1
        

print(C)

你可以在没有任何模块的情况下尝试这个。

您可以指定数组，无论如何都可以。我就是这样测试的

Answer 5

更多变化和基准：

Best 3 of 20 rounds:
 885 ns   888 ns   888 ns  [*map(concat, A, cycle(B))]
 955 ns   959 ns   961 ns  list(map(concat, A, cycle(B)))
 969 ns   976 ns   976 ns  list(map(add, A, cycle(B)))
1168 ns  1173 ns  1174 ns  list(starmap(concat, zip(A, cycle(B))))
1230 ns  1235 ns  1239 ns  list(map(''.join, zip(A, cycle(B))))
1232 ns  1236 ns  1249 ns  [a + b for a, b in zip(A, cycle(B))]
1335 ns  1344 ns  1346 ns  [f'{a}{b}' for a, b in zip(A, cycle(B))]
1359 ns  1362 ns  1366 ns  [a + B[i%2] for i, a in enumerate(A)]
1418 ns  1421 ns  1423 ns  m = len(B); [a + B[i%m] for i, a in enumerate(A)]
1492 ns  1509 ns  1510 ns  [A[i] + B[i%2] for i in range(len(A))]
1964 ns  1978 ns  1978 ns  list(map('{}{}'.format, A, cycle(B)))

A 乘以 1000（即列表长 1000 倍）：

Best 3 of 20 rounds:
 654 us   655 us   658 us  [*map(concat, A, cycle(B))]
 657 us   657 us   657 us  list(map(concat, A, cycle(B)))
 674 us   676 us   676 us  list(map(add, A, cycle(B)))
 727 us   737 us   738 us  list(starmap(concat, zip(A, cycle(B))))
 743 us   746 us   747 us  list(map(''.join, zip(A, cycle(B))))
 837 us   841 us   841 us  [a + b for a, b in zip(A, cycle(B))]
 943 us   945 us   946 us  [f'{a}{b}' for a, b in zip(A, cycle(B))]
1279 us  1288 us  1290 us  m = len(B); [a + B[i%m] for i, a in enumerate(A)]
1280 us  1280 us  1281 us  [a + B[i%2] for i, a in enumerate(A)]
1317 us  1319 us  1323 us  [A[i] + B[i%2] for i in range(len(A))]
1626 us  1633 us  1638 us  list(map('{}{}'.format, A, cycle(B)))

代码（Try it online!）：

from timeit import repeat
from random import shuffle
from bisect import insort

setup = '''
from itertools import cycle, starmap
from operator import concat, add
A = ['A','B','C','D','E','F','G','H']
B = ['a','b']
'''

solutions = [
    'list(map(add, A, cycle(B)))',
    'list(map(concat, A, cycle(B)))',
    '[*map(concat, A, cycle(B))]',
    '[a + B[i%2] for i, a in enumerate(A)]',
    'm = len(B); [a + B[i%m] for i, a in enumerate(A)]',
    "list(map(''.join, zip(A, cycle(B))))",
    "list(map('{}{}'.format, A, cycle(B)))",
    '[a + b for a, b in zip(A, cycle(B))]',
    "[f'{a}{b}' for a, b in zip(A, cycle(B))]",
    'list(starmap(concat, zip(A, cycle(B))))',
    '[A[i] + B[i%2] for i in range(len(A))]',
]

exec(setup)
for sol in solutions:
    try:
        print(eval(sol) == ['Aa','Bb','Ca','Db','Ea','Fb','Ga','Hb'])
    except SyntaxError:
        pass

tss = {sol: [] for sol in solutions}
for i in range(20):
    print(f'Best 3 of {i+1} rounds:')
    shuffle(solutions)
    for sol in solutions:
        number = 10 ** 4
        t = min(repeat(sol, setup, number=number)) / number
        insort(tss[sol], t)
    for sol in sorted(tss, key=tss.get):
        print(*('%4d ns ' % (t * 1e9) for t in tss[sol][:3]), sol)
    print()

Answer 6

你也可以试试这个：

A = ['A','B','C','D','E','F','G','H']
B= ['a','b']
LB = len(B)
x = 0

for i in A :
    if x>=LB :
        x=0
    print(i+B[x])
    x+=1

python 当外循环索引等于内循环长度时嵌套for循环你重新开始

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