Itertools合并两个列表以获得所有可能的组合

Question

我有两个列表：a和b。你知道吗

a是包含三个或更多字符串的列表，而b是分隔符列表。你知道吗

我需要生成a的所有可能组合，然后将结果与b的所有可能组合“合并”（请参见示例以获得更好的理解）。你知道吗

我最终使用了以下代码：

from itertools import permutations, combinations, product

a = ["filename", "timestamp", "custom"]
b = ["_", "-", ".", ""]

output = []

for com in combinations(b, len(a) - 1):
    for per in product(com, repeat=len(a) - 1):
        for ear_per in permutations(a):
            out = ''.join(map(''.join, zip(list(ear_per[:-1]), per))) + list(ear_per)[-1]
            output.append(out)

# For some reason the algorithm is generating duplicates
output = list(dict.fromkeys(output))

for o in output:
    print o

这是一个输出示例（这是正确的，这是我在本例中需要的）：

timestamp.customfilename
filenamecustom.timestamp
custom_filenametimestamp
timestamp_custom_filename
timestamp-filename.custom
custom_filename-timestamp
filename.timestamp-custom
. . .
filename.custom.timestamp
filename-customtimestamp
custom-timestamp_filename
filename_custom-timestamp
filename.timestampcustom
timestampcustom-filename
custom-timestamp.filename
filenamecustom_timestamp
timestamp.custom_filename
custom.timestampfilename
timestampfilename.custom
customfilename_timestamp
filenametimestamp-custom
custom-filenametimestamp
timestampfilename-custom
timestamp-custom-filename
custom.filenametimestamp
customfilenametimestamp
timestampfilename_custom
custom_filename.timestamp
custom-timestamp-filename
custom-timestampfilename
filename_timestamp.custom
. . .
filename.custom-timestamp
timestamp_filenamecustom
custom_timestampfilename
timestamp.custom.filename
timestamp.filename-custom
filename-custom-timestamp
customfilename.timestamp
filename_timestamp_custom
timestamp_filename.custom
customtimestampfilename
filenamecustomtimestamp
custom.timestamp_filename
filename_customtimestamp
. . .
timestamp-customfilename
filename_custom.timestamp

此算法有两个主要问题：

1. 它会生成一些重复的行，所以我总是需要删除它们（在较大的数据集上速度较慢）

2. if len(a) > len(b) + 2脚本无法启动。在这种情况下，我需要重复分隔符以覆盖a中包含的单词之间的len(a) - 1可用空格。

Answer 1

这可能是一个可行的解决办法。它将a的(3*2 = 6)的排列与b的(2 at a time here, 4*4 == 16)的product交错，得到总共6 * 16 == 96个结果。你知道吗

from itertools import permutations, chain, zip_longest, product

a = ["filename", "timestamp", "custom"]
b = ["_", "-", ".", ""]

i=0
for perm in permutations(a):
    for prod in product(b,repeat=len(a)-1):
        tpls = list(chain.from_iterable(zip_longest(perm, prod, fillvalue='')))
        print(''.join(tpls))
        i += 1
print(i)