创建可能案例的简单方法

Question

我有数据列表，例如

a = [1,2,3,4]
b = ["a","b","c","d","e"]
c = ["001","002","003"]

我想创建一个新的另一个列表，该列表是由 a、b、c 的所有可能情况混合而成的 是否

d = ["1a001","1a002","1a003",...,"4e003"]

有任何模块或方法可以生成 d 而无需编写许多 for 循环？

Answer 1

[''.join(str(y) for y in x) for x in itertools.product(a, b, c)]

Answer 2

如果您也将 a 转换为 str 列表，那么这样做会更简单。
并节省了对 b 和 c 的每个元素不必要的调用 str()

>>> from itertools import product
>>> a = [1,2,3,4]
>>> b = ["a","b","c","d","e"]
>>> c = ["001","002","003"]
>>> a = map(str,a)                   # so a=['1', '2', '3', '4']
>>> map(''.join, product(a, b, c))
['1a001', '1a002', '1a003', '1b001', '1b002', '1b003', '1c001', '1c002', '1c003', '1d001', '1d002', '1d003', '1e001', '1e002', '1e003', '2a001', '2a002', '2a003', '2b001', '2b002', '2b003', '2c001', '2c002', '2c003', '2d001', '2d002', '2d003', '2e001', '2e002', '2e003', '3a001', '3a002', '3a003', '3b001', '3b002', '3b003', '3c001', '3c002', '3c003', '3d001', '3d002', '3d003', '3e001', '3e002', '3e003', '4a001', '4a002', '4a003', '4b001', '4b002', '4b003', '4c001', '4c002', '4c003', '4d001', '4d002', '4d003', '4e001', '4e002', '4e003']

这是时间比较

timeit astr=map(str,a);map(''.join, product(astr, b, c))
10000 loops, best of 3: 43 us per loop

timeit [''.join(str(y) for y in x) for x in product(a, b, c)]
1000 loops, best of 3: 399 us per loop

Answer 3

map(''.join, itertools.product(map(str, a), b, c))

Answer 4

这个怎么样？使用地图

print [''.join(map(str,y)) for y in itertools.product(a,b,c)]