Question

我有两个清单：

a= [['A', 'B', 'C', 3], ['P', 'Q', 'R', 4]]

b=[['K',1,1,1,1,1], ['L',1,1,1,1,1], ['M', 1,1,0,1,1], ['J', 0,0,0,0,0], ['A', 0,0,0,1,1], ['P',0,1,0,1,1 ]]

我希望输出如下：

Output=[['A', 0,0,0,1,1], ['P',0,1,0,1,1 ]]

我正在尝试使用a [idx] [0]搜索a中的a。然后我想收集这些项目，并希望像上面的输出。

我的代码如下：

Output=[]
for idx in range(len(Test)):
    a_idx = [y[0] for y in b].index(a[idx][0])
    a_in_b = b[a_idx]
    Output.append(a_in_b[:])

print Output

这不能给我想要的输出。有人可以帮忙吗？

Answer 1

首先，将b转换为字典：

b=[['K',1,1,1,1,1], ['L',1,1,1,1,1], ['M', 1,1,0,1,1], ['J', 0,0,0,0,0], ['A', 0,0,0,1,1], ['P',0,1,0,1,1 ]

d = dict((i[0], i[1:]) for i in b)

# d is now:
{'A': [0, 0, 0, 1, 1],
 'J': [0, 0, 0, 0, 0],
 'K': [1, 1, 1, 1, 1],
 'L': [1, 1, 1, 1, 1],
 'M': [1, 1, 0, 1, 1],
 'P': [0, 1, 0, 1, 1]}

然后将d映射到a：

Output = [ i[:1] + d[i[0]] for i in a]

# Output is now: [['A', 0, 0, 0, 1, 1], ['P', 0, 1, 0, 1, 1]]

Answer 2

虽然eumiro的答案更好，但您要求使用索引的版本。我的版本似乎始终如一：

a= [['A', 'B', 'C', 3], ['P', 'Q', 'R', 4]]
b=[['K',1,1,1,1,1], ['L',1,1,1,1,1], ['M', 1,1,0,1,1], ['J', 0,0,0,0,0], ['A', 0,0,0,1,1], ['P',0,1,0,1,1 ]]
src = [y[0] for y in b]; # I moved this out here so that it is only calculated once
Output = []
for i in range(len(a)):  # You have Test here instead??? Not sure why
     ai  = src.index( a[ i ][ 0 ] )
     Output.append( b[ ai ][:] )

搜索操作有问题

2 个答案: