我有兩個串列（X和Y）。我想先X用 的所有元素縮放第一個元素，Y然后再移動到第二個元素，X然后再用 的所有元素縮放Y。我怎樣才能做到這一點？理想情況下，當 X 的新元素開始時，我還想將它附加到不同的串列 ( L)，但我不確定這是否可能。
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)
首選結果：
# 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
uj5u.com熱心網友回復：
這似乎遵循您的模式：
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)
輸出：
[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]
根據 OP 對索引進行分組的評論：
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)
輸出：
[[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]]
uj5u.com熱心網友回復：
首先，您可以通過直接訪問專案來簡單地回圈，而無需索引。然后您可以將內部回圈轉換為理解串列以使其更緊湊：
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]
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