英文:
Generating a list based on other lists in Python
问题
以下是翻译好的代码部分：
我有两个列表```A2```和```J```。我正在执行一个操作，每当```J[0][i]=0```时，```A3==[0]```。我呈现当前和期望的输出。```pythonA2 = [[2, 3, 5], [3, 4, 6], [0, 3, 5], [0, 1, 2, 4, 5, 6], [1, 3, 6], [0, 2, 3, 7, 8, 10],[1, 3, 4, 8, 9, 11], [5, 8, 10], [5, 6, 7, 9, 10, 11], [6, 8, 11], [5, 7, 8], [6, 8, 9]]J=[[0, 2, 0, 6, 7, 9, 10]]A3=[]for i in range(0,len(J[0])):if (J[0][i]==0):A3==[0]else:A33=A2[J[0][i]]A3.append(A33)print("A3 =",A3)
当前输出是：
A3 = [[0, 3, 5], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
期望输出是：
A3 = [[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
请注意，期望输出中每个子列表都包含一个0，而当前输出中没有。
英文:
I have two lists A2 and J.
I am performing an operation such that whenever J[0][i]=0, A3==[0]. I present the current and expected output.
A2 = [[2, 3, 5], [3, 4, 6], [0, 3, 5], [0, 1, 2, 4, 5, 6], [1, 3, 6], [0, 2, 3, 7, 8, 10],[1, 3, 4, 8, 9, 11], [5, 8, 10], [5, 6, 7, 9, 10, 11], [6, 8, 11], [5, 7, 8], [6, 8, 9]]J=[[0, 2, 0, 6, 7, 9, 10]]A3=[]for i in range(0,len(J[0])):if (J[0][i]==0):A3==[0]else:A33=A2[J[0][i]]A3.append(A33)print("A3 =",A3)
The current output is :
A3 = [[0, 3, 5], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
The expected output is :
A3 = [[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
答案1
得分: 2
A2 = [[2, 3, 5], [3, 4, 6], [0, 3, 5], [0, 1, 2, 4, 5, 6], [1, 3, 6], [0, 2, 3, 7, 8, 10],[1, 3, 4, 8, 9, 11], [5, 8, 10], [5, 6, 7, 9, 10, 11], [6, 8, 11], [5, 7, 8], [6, 8, 9]]J = [[0, 2, 0, 6, 7, 9, 10]]A3 = []for i in range(0, len(J[0])):if (J[0][i] == 0):A3.append([0]) # 修改这里else:A33 = A2[J[0][i]]A3.append(A33)print("A3 =", A3)# 输出A3 = [[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
注意：在 A3==[0] 的情况下，请改成 A3.append([0])，因为 A3==[0] 是在比较 A3 和 0，不会对 A3 进行任何更改。
英文:
A2 = [[2, 3, 5], [3, 4, 6], [0, 3, 5], [0, 1, 2, 4, 5, 6], [1, 3, 6], [0, 2, 3, 7, 8, 10],[1, 3, 4, 8, 9, 11], [5, 8, 10], [5, 6, 7, 9, 10, 11], [6, 8, 11], [5, 7, 8], [6, 8, 9]]J=[[0, 2, 0, 6, 7, 9, 10]]A3=[]for i in range(0,len(J[0])):if (J[0][i]==0):A3.append([0]) # change hereelse:A33=A2[J[0][i]]A3.append(A33)print("A3 =",A3)#outputA3 = [[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
Instead of A3==[0] please do A3.append([0]) as A3==[0] is comparing A3 with 0 and it is doing no change to A3
答案2
得分: 0
在你的代码中进行以下更改：
A3 = []for i in J[0]:if i == 0:A3.extend([[0]])else:A3.extend([A2[i]])
现在A3的值为：
[[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
解释：
在你之前的代码中，A3 == 0不会将[0]插入到列表中，而是返回一个布尔值。由于你想要将0插入作为单个元素列表，你可以使用extend函数，它允许你附加带有多个元素的列表。
英文:
Make the following changes in your code:
A3 = []for i in J[0]:if i == 0:A3.extend([[0]])else:A3.extend([A2[i]])
Now the value of A3 is:
[[0], [0, 3, 5], [0], [1, 3, 4, 8, 9, 11], [5, 8, 10], [6, 8, 11], [5, 7, 8]]
Explaination:
In your previous code A3==0 does not insert [0] to your list instead it returs a bool. Since you want to insert 0 as a single element list you can use extend function which lets you append a list with multiple elements.
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