創建可能組合列表的列表

Question

我有一個列表，其中包含以下項目：

initl = [1, 2, 3, 4]

和一個stepsize ，即stepsize = 0.05 。 列表中的每個值都可以按stepsize向上或向下更改。 我想要創建的是一個列表列表，其中包含向上、向下或初始值的所有組合，例如：

result_list = [[1, 2, 3, 4], [1.05, 2, 3, 4], [0.95, 2, 3, 4], [1, 2.05, 3, 4], ...] 

列表中的組合順序無關緊要。

我想出了這個：

import itertools


initl = [1, 2, 3, 4]
stepsize = 0.05

lu = [i + stepsize for i in initl]
ld = [i - stepsize for i in initl]

l_map = list(itertools.product(range(3), repeat=4))

result_list = []

for i in l_map:
    l_new = []
    for pos, j in enumerate(i):
        if j == 0:
            l_new.append(ld[pos])
        elif j == 1:
            l_new.append(initl[pos])
        else:
            l_new.append(lu[pos])
   result_list.append(l_new)

這會產生所需的 output。 列表長度：3 ^ 4 = 81。但我想知道是否有更好的方法。 特別是嵌套的 for 循環對我來說似乎很笨重。 任何幫助表示贊賞。

Answer 1

你有正確的想法，基本上實現numpy.choose 。選擇你自己，對於一個 3 選擇的情況。

您可以直接在包含每個值的所有選項的元組的列表上簡化使用itertools.product ，如下所示：

import itertools

initl = [1, 2, 3, 4]
stepsize = 0.05

prod_seed = [(i, i+stepsize, i-stepsize) for i in initl]
result_list = list(itertools.product(*prod_seed))

print(result_list)
print(len(result_list))

Output：

[(1, 2, 3, 4), (1, 2, 3, 4.05), (1, 2, 3, 3.95), ....]
81

Answer 2

嘗試：

def getComb(arr, step, res=[]):
    if(len(arr)==1):
        for el in [-step, 0, step]:
            yield res+[arr[0]+el]
    else:
        for el in [-step, 0, step]:
            yield from getComb(arr[1:], step, res+[arr[0]+el])

對於您的輸入數據輸出：

>>> for el in getComb(initl, stepsize): print(el)

[0.95, 1.95, 2.95, 3.95]
[0.95, 1.95, 2.95, 4]
[0.95, 1.95, 2.95, 4.05]
[0.95, 1.95, 3, 3.95]
[0.95, 1.95, 3, 4]
...
[1.05, 2.05, 3, 4]
[1.05, 2.05, 3, 4.05]
[1.05, 2.05, 3.05, 3.95]
[1.05, 2.05, 3.05, 4]
[1.05, 2.05, 3.05, 4.05]

Answer 3

您可以嘗試如下的列表理解， zip使用initl對product s 執行操作：

>>> from itertools import product
>>> initl = [1, 2, 3, 4]
>>> step = [-0.05, 0, 0.05]
>>> [[x+d for x, d in zip(initl, p)] for p in product(step, repeat=len(initl))]
[[0.95, 1.95, 2.95, 3.95],
 [0.95, 1.95, 2.95, 4],
 ...
 [1.05, 2.05, 3.05, 4.05]]
>>> len(_)
81

Answer 4

您快到了。 您需要做的就是轉置由 3 個（4 個長度）列表組成的矩陣，以便itertools.product可以從[elem[i] - stepsize, elem[i], elem[i] + stepsize]中選擇每個元素. 這是由zip function 完成的。

 >>> import itertools as it >>> >>> initl = [1, 2, 3, 4] >>> stepsize = 0.05 >>> >>> lo = [elem - stepsize for elem in initl] >>> hi = [elem + stepsize for elem in initl] >>> lo, hi ([0.95, 1.95, 2.95, 3.95], [1.05, 2.05, 3.05, 4.05]) >>> >>> list(it.product(*zip(lo, initl, hi))) [(0.95, 1.95, 2.95, 3.95), (0.95, 1.95, 2.95, 4), (0.95, 1.95, 2.95, 4.05), (0.95, 1.95, 3, 3.95), (0.95, 1.95, 3, 4), (0.95, 1.95, 3, 4.05), (0.95, 1.95, 3.05, 3.95), (0.95, 1.95, 3.05, 4), (0.95, 1.95, 3.05, 4.05), (0.95, 2, 2.95, 3.95), (0.95, 2, 2.95, 4), (0.95, 2, 2.95, 4.05), (0.95, 2, 3, 3.95), (0.95, 2, 3, 4), (0.95, 2, 3, 4.05), (0.95, 2, 3.05, 3.95), (0.95, 2, 3.05, 4), (0.95, 2, 3.05, 4.05), (0.95, 2.05, 2.95, 3.95), (0.95, 2.05, 2.95, 4), (0.95, 2.05, 2.95, 4.05), (0.95, 2.05, 3, 3.95), (0.95, 2.05, 3, 4), (0.95, 2.05, 3, 4.05), (0.95, 2.05, 3.05, 3.95), (0.95, 2.05, 3.05, 4), (0.95, 2.05, 3.05, 4.05), (1, 1.95, 2.95, 3.95), (1, 1.95, 2.95, 4), (1, 1.95, 2.95, 4.05), (1, 1.95, 3, 3.95), (1, 1.95, 3, 4), (1, 1.95, 3, 4.05), (1, 1.95, 3.05, 3.95), (1, 1.95, 3.05, 4), (1, 1.95, 3.05, 4.05), (1, 2, 2.95, 3.95), (1, 2, 2.95, 4), (1, 2, 2.95, 4.05), (1, 2, 3, 3.95), (1, 2, 3, 4), (1, 2, 3, 4.05), (1, 2, 3.05, 3.95), (1, 2, 3.05, 4), (1, 2, 3.05, 4.05), (1, 2.05, 2.95, 3.95), (1, 2.05, 2.95, 4), (1, 2.05, 2.95, 4.05), (1, 2.05, 3, 3.95), (1, 2.05, 3, 4), (1, 2.05, 3, 4.05), (1, 2.05, 3.05, 3.95), (1, 2.05, 3.05, 4), (1, 2.05, 3.05, 4.05), (1.05, 1.95, 2.95, 3.95), (1.05, 1.95, 2.95, 4), (1.05, 1.95, 2.95, 4.05), (1.05, 1.95, 3, 3.95), (1.05, 1.95, 3, 4), (1.05, 1.95, 3, 4.05), (1.05, 1.95, 3.05, 3.95), (1.05, 1.95, 3.05, 4), (1.05, 1.95, 3.05, 4.05), (1.05, 2, 2.95, 3.95), (1.05, 2, 2.95, 4), (1.05, 2, 2.95, 4.05), (1.05, 2, 3, 3.95), (1.05, 2, 3, 4), (1.05, 2, 3, 4.05), (1.05, 2, 3.05, 3.95), (1.05, 2, 3.05, 4), (1.05, 2, 3.05, 4.05), (1.05, 2.05, 2.95, 3.95), (1.05, 2.05, 2.95, 4), (1.05, 2.05, 2.95, 4.05), (1.05, 2.05, 3, 3.95), (1.05, 2.05, 3, 4), (1.05, 2.05, 3, 4.05), (1.05, 2.05, 3.05, 3.95), (1.05, 2.05, 3.05, 4), (1.05, 2.05, 3.05, 4.05)]