迭代乘以 python 中的两个列表的元素

Question

我有两个列表（X 和 Y）。我想先用 Y 的所有元素缩放 X 的第一个元素，然后再移动到 X 的第二个元素，然后再次按 [=14= 的所有元素缩放].我怎样才能做到这一点？理想情况下，我还想在启动 X 的新元素时将其附加到不同的列表 (L)，但我不确定这是否可能。

Y = [2, 3, 4, 5]
X = [1, 2, 3, 4]
L = []
for i in range(len(X)):
    for j in range(len(Y)):
        X_scale = X[i] * Y[j]
        L.append(X_scale)

首选结果：

# First element in X
X_scale = [2, 2, 3, 4]
X_scale = [3, 2, 3, 4]
X_scale = [4, 2, 3, 4]
X_scale = [5, 2, 3, 4]

# Second element in X
X_scale = [1, 4, 3, 4]
X_scale = [1, 6, 3, 4]
#etc

Answer 1

首先，您可以通过直接访问项目来简单地循环，而无需索引。然后你可以将内层循环转化为一个理解列表，使其更紧凑：

Y = [2, 3, 4, 5]
X = [1, 2, 3, 4]
L = []
for x_item in X:
    L += [x_item * y_item for y_item in Y]

Answer 2

这似乎符合您的模式：

Y = [2, 3, 4, 5]
X = [1, 2, 3, 4]
L = []
for i,x in enumerate(X):
    for y in Y:
        X_scale = X.copy()
        X_scale[i] = x * y
        L.append(X_scale)

for row in L:
    print(row)

输出：

[2, 2, 3, 4]
[3, 2, 3, 4]
[4, 2, 3, 4]
[5, 2, 3, 4]
[1, 4, 3, 4]
[1, 6, 3, 4]
[1, 8, 3, 4]
[1, 10, 3, 4]
[1, 2, 6, 4]
[1, 2, 9, 4]
[1, 2, 12, 4]
[1, 2, 15, 4]
[1, 2, 3, 8]
[1, 2, 3, 12]
[1, 2, 3, 16]
[1, 2, 3, 20]

根据 OP 对索引进行分组的评论：

Y = [2, 3, 4, 5]
X = [1, 2, 3, 4]
L = []
for i,x in enumerate(X):
    L2 = []
    for y in Y:
        X_scale = X.copy()
        X_scale[i] = x * y
        L2.append(X_scale)
    L.append(L2)

for row in L:
    print(row)

输出：

[[2, 2, 3, 4], [3, 2, 3, 4], [4, 2, 3, 4], [5, 2, 3, 4]]
[[1, 4, 3, 4], [1, 6, 3, 4], [1, 8, 3, 4], [1, 10, 3, 4]]
[[1, 2, 6, 4], [1, 2, 9, 4], [1, 2, 12, 4], [1, 2, 15, 4]]
[[1, 2, 3, 8], [1, 2, 3, 12], [1, 2, 3, 16], [1, 2, 3, 20]]

迭代乘以 python 中的两个列表的元素

Iteratively multiply elements of two lists in python

python

arrays

list

multiplication