我如何在 Python 中執(zhí)行此操作？輸入：num = [1, 2, 3]alpha = ['a', 'b', 'c']輸出：[1, a] [1, a, b] [1, a, b, c] [1, b, c] [1, c][1, 2, a] [1, 2, a, b] ... [1, 2, c]......[3, a] [3, a, b] ... [3, c]約束：The output expect all possible consecutive combinationse.g. ['a'] ['a','b'] are desired while ['a','c'] is not.我嘗試了 4 個(gè) for 循環(huán)/4 個(gè) while 循環(huán)的嵌套。使用這種深度的循環(huán)是否常見？
3 回答
千萬里不及你
TA貢獻(xiàn)1784條經(jīng)驗(yàn) 獲得超9個(gè)贊
以下僅使用兩個(gè)嵌套的 for 循環(huán)
from itertools import product as prod
def consecutive_combos(a):
" Generates consecutive ombinations of items in list "
return [a[i:j] for i in range(len(a)) for j in range(i+1, len(a)+1)]
num = [1, 2, 3]
alpha = ['a', 'b', 'c']
# Generates product of sequences
result = [x + y for x, y in prod(consecutive_combos(num), consecutive_combos(alpha))]
print(result)
輸出
[[1, 'a'], [1, 'a', 'b'], [1, 'a', 'b', 'c'], [1, 'b'], [1, 'b', 'c'], [1, 'c'], [1, 2, 'a'], [1, 2, 'a', 'b'], [1, 2, 'a', 'b', 'c'], [1, 2, 'b'], [1, 2, 'b', 'c'], [1, 2, 'c'], [1, 2, 3, 'a'], [1, 2, 3, 'a', 'b'], [1, 2, 3, 'a', 'b', 'c'], [1, 2, 3, 'b'], [1, 2, 3, 'b', 'c'], [1, 2, 3, 'c'], [2, 'a'], [2, 'a', 'b'], [2, 'a', 'b', 'c'], [2, 'b'], [2, 'b', 'c'], [2, 'c'], [2, 3, 'a'], [2, 3, 'a', 'b'], [2, 3, 'a', 'b', 'c'], [2, 3, 'b'], [2, 3, 'b', 'c'], [2, 3, 'c'], [3, 'a'], [3, 'a', 'b'], [3, 'a', 'b', 'c'], [3, 'b'], [3, 'b', 'c'], [3, 'c']]
翻閱古今
TA貢獻(xiàn)1780條經(jīng)驗(yàn) 獲得超5個(gè)贊
您可以使用模塊itertoolsanscombinations
import itertools as it
l1 = []
# get all combinations of num
for i in range(1, len(num)+1):
l1.extend([* it.combinations(num, i)])
# get all combinations of alpha
for j in range(1, len(alpha)+1):
l2.extend([* it.combinations(alpha, j)])
# list comprehension to combine elements from the two lists
comb = [e1+e2 for e1 in l1 for e2 in l2]
通過這樣做，您將獲得一個(gè)元組列表。
更新：
為了考慮約束：
輸出期望所有可能的連續(xù)組合
例如 [a] [a,b] 是需要的，而 [a,c] 不是。
num = [1, 2, 3]
alpha = ['a', 'b', 'c']
l1 = [num[i:j+1] for i in range(len(num)) for j in range(i, len(num))]
l2 = [alpha[i:j+1] for i in range(len(alpha)) for j in range(i, len(alpha))]
result = [e1+e2 for e1 in l1 for e2 in l2]
print(*result, sep="\n")
[1, 'a']
[1, 'a', 'b']
[1, 'a', 'b', 'c']
[1, 'b']
[1, 'b', 'c']
[1, 'c']
[1, 2, 'a']
[1, 2, 'a', 'b']
[1, 2, 'a', 'b', 'c']
[1, 2, 'b']
[1, 2, 'b', 'c']
[1, 2, 'c']
[1, 2, 3, 'a']
[1, 2, 3, 'a', 'b']
[1, 2, 3, 'a', 'b', 'c']
[1, 2, 3, 'b']
[1, 2, 3, 'b', 'c']
[1, 2, 3, 'c']
[2, 'a']
[2, 'a', 'b']
[2, 'a', 'b', 'c']
[2, 'b']
[2, 'b', 'c']
[2, 'c']
[2, 3, 'a']
[2, 3, 'a', 'b']
[2, 3, 'a', 'b', 'c']
[2, 3, 'b']
[2, 3, 'b', 'c']
[2, 3, 'c']
[3, 'a']
[3, 'a', 'b']
[3, 'a', 'b', 'c']
[3, 'b']
[3, 'b', 'c']
[3, 'c']
largeQ
TA貢獻(xiàn)2039條經(jīng)驗(yàn) 獲得超8個(gè)贊
我希望它會(huì)有所幫助
from itertools import combinations
num = [1, 2, 3]
alpha = ["a", "b", "c"]
big_list=num+alpha
comb += list(combinations(big_list,2) )
comb += list(combinations(big_list,3) )
for i in list(comb):
print (i)
輸出：
(1, 2) (1, 3) (1, 'a') (1, 'b') (1, 'c') (2, 3) (2, 'a') (2, 'b') (2, 'c') (3, 'a') (3, 'b') (3, 'c') ('a', 'b') ('a', 'c') ('b' , 'c') (1, 2, 3) (1, 2, 'a') (1, 2, 'b') (1, 2, 'c') (1, 3, 'a') (1 , 3, 'b') (1, 3, 'c') (1, 'a', 'b') (1, 'a', 'c') (1, 'b', 'c') ( 2, 3, 'a') (2, 3, 'b') (2, 3, 'c') (2, 'a', 'b') (2, 'a', 'c') (2 , 'b', 'c') (3, 'a', 'b') (3, 'a', 'c') (3, 'b', 'c') ('a', 'b' , 'c') [0.50秒內(nèi)完成]
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