英文:
Creating multiple sublists by performing an operation in Python
问题
当前的输出是：
[0.0, 0.9979508721068377, 0.9961113206802571, 0.9979508721068377, 0.0, 0.0]
期望的输出是：
[[0.0, 0.9979508721068377, 0.9961113206802571], [0.9979508721068377, 0.0, 0.0]]
英文:
I have two lists A2 and Cb_new. I am performing an operating as shown below. But I want to create multiple sublists instead of one single list. I present the current and expected outputs.
A2=[[2, 3, 5], [3, 4, 6]]Cb_new=[[1.0, 0.0, 0.0, 0.9979508721068377, 0.0, 0.9961113206802571, 0.0, 0.0, 0.996111320680257, 0.0, 0.0]]Cb=[]for j in range(0,len(A2)):for i in range(0,len(A2[j])):Cb1=Cb_new[0][A2[j][i]]Cb.append(Cb1)print(Cb)
The current output is
[0.0, 0.9979508721068377, 0.9961113206802571, 0.9979508721068377, 0.0, 0.0]
The expected output is
[[0.0, 0.9979508721068377, 0.9961113206802571], [0.9979508721068377, 0.0, 0.0]]
答案1
得分: 1
您需要使用临时列表来实现这种行为。在您的代码中，Cb1始终是Cb_new列表中的一个元素。相反，您应该将它附加到一个在外部for循环的每次迭代中重置的列表中。for j in range(0,len(A2)):Cb_interm = []for i in range(0,len(A2[j])):Cb1=Cb_new[0][A2[j][i]]Cb_interm.append(Cb1)Cb.append(Cb_interm)print(Cb)输出结果:[[0.0, 0.9979508721068377, 0.9961113206802571], [0.9979508721068377, 0.0, 0.0]]
英文:
You will need to use a temporary list to achieve this behavior. In your code Cb1 is always an element from Cb_new list. Instead, you should append it to a list that you reset at every iteration of outer for loop.
for j in range(0,len(A2)):Cb_interm = []for i in range(0,len(A2[j])):Cb1=Cb_new[0][A2[j][i]]Cb_interm.append(Cb1)Cb.append(Cb_interm)print(Cb)
The output:
[[0.0, 0.9979508721068377, 0.9961113206802571], [0.9979508721068377, 0.0, 0.0]]
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