Question

我试图想出一个函数，它接受一个输入x并将一个大的列表与元素x * x的数量分成x个较小的列表，每个列表中包含x个元素例如：

big_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

def split_list (x):
big_list = pairs (x)
small_list = [big_list[0:x] for x in range (x)]

我的输出必须是：

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

但我没有得到它，你推荐什么？

Answer 1

你可以试试这个：

big_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

def split_list (x):
   return [big_list[i:i+x] for i in range(0, len(big_list), x)]

print(split_list(4))

输出：

[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

Answer 2

首先，您希望使用平方根来获取大小，因此，对于大小为n的列表，您将拥有m x m矩阵m = n ** 0.5。让我们首先定义您的功能：

def square(array):
    n = len(array)
    m = int(n ** 0.5)

由于列表大小必须是整数，因此我们必须在结果上调用int。接下来，我们希望每次从0开始计数i至n m，从m开始执行i元素：

def square(array):
    n = len(array)
    m = int(n ** 0.5)
    result = []
    for i in range(0, n, m):
        result.append(array[i:i + m])
    return result

就是这样。

或者，s=lambda a:(lambda m:[a[i*m:i*m+m]for i in range(m)])(int(len(a)**.5))

Answer 3

我一直都在使用这段代码。

def chunkify(items, chunk_len):
    return [items[i:i+chunk_len] for i in range(0,len(items),chunk_len)]

Answer 4

这是一个解决方案：

def split_list(big_list, x):
    list_size = len(big_list)
    splits = int(list_size / x)

    return [big_list[k*x:k*x+x] for k in range(splits)]

big_list = [i+1 for i in range(16)]
print(big_list)
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

small_list = split_list(big_list, 4)
print(small_list)
# [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]


small_list = split_list(big_list, 2)
print(small_list)
# [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10], [11, 12], [13, 14], [15, 16]]

small_list = split_list(big_list, 3)
print(small_list)
# [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12], [13, 14, 15]]

Answer 5

为什么不使用numpy：

>>> import numpy as np
>>> big_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
>>> a=np.array(big_list)
>>> a
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16])
>>> n=int(len(big_list)**0.5)
>>> a.reshape(n,n)
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12],
       [13, 14, 15, 16]])

或只是：

>>> big_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]
>>> n=int(len(big_list)**0.5)
>>> new_list=[big_list[i:i+n] for i in range(0, len(big_list), n)]
>>> new_list
[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]

将包含n * n个元素的列表拆分为n个列表，每个列表中包含n个元素

5 个答案: