我如何在 Python 中執行此操作？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 for 循環/4 個 while 循環的嵌套。使用這種深度的循環是否常見？
3 回答
千萬里不及你
TA貢獻1784條經驗 獲得超9個贊
以下僅使用兩個嵌套的 for 循環
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1780條經驗 獲得超5個贊
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，您將獲得一個元組列表。
更新：
為了考慮約束：
輸出期望所有可能的連續組合
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2039條經驗 獲得超8個贊
我希望它會有所幫助
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秒內完成]
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