Question

>>>lis = df['col'].values.tolist()    

>>>lis = [[(('A', 'WE'), ('1,21', 'rr'), ('io', 'mp'))], 
     [(('B', 'WE'), ('5', 'rr'), ('io', 'mp'))],
     [(('A', 'WE'), ('3', 'rr'), ('io', 'mp')),
      (('C', 'WE'), ('0', 'rr'), ('io', 'mp'))],
     ....
     [(('D', 'WE'), ('6', 'rr'), ('io', 'mp'))],
     [(('A', 'WE'), ('9.0', 'rr'), ('io', 'mp'))]]

如何只获取每个元组的第一个元素并将其重新格式化为：

 [[A, 1,21, io],
 [B, 5, io],
 [A, 3, io],
 [C, 0, io],
 ....
 [D, 6, io],
 [A, 9.0', io]]

我已经准备好了：

[' '.join(map(str,lis[0][0])) for x in lis]

和

[' '.join(map(str,lis[0][:1])) for x in lis]

和

' '.join(map(str, lis))

和

new_lis, _ = zip(*lis[0][0])    
return ' '.join(new_lis)

更新

pandas列如下所示：

   Col
0 [(('A', 'WE'), ('1,21', 'rr'), ('io', 'mp'))] 
1 [(('B', 'WE'), ('5', 'rr'), ('io', 'mp'))]
2 [(('A', 'WE'), ('3', 'rr'), ('io', 'mp'))
3 [(('C', 'WE'), ('0', 'rr'), ('io', 'mp'))]
   ....
n   [(('D', 'WE'), ('6', 'rr'), ('io', 'mp'))]
n-1 [(('A', 'WE'), ('9.0', 'rr'), ('io', 'mp'))]

Answer 1

lis = [[(('A', 'WE'), ('1,21', 'rr'), ('io', 'mp'))], 
     [(('B', 'WE'), ('5', 'rr'), ('io', 'mp'))],
     [(('A', 'WE'), ('3', 'rr'), ('io', 'mp')),
      (('C', 'WE'), ('0', 'rr'), ('io', 'mp'))],
     [(('D', 'WE'), ('6', 'rr'), ('io', 'mp'))],
     [(('A', 'WE'), ('9.0', 'rr'), ('io', 'mp'))]]

print [list(next(zip(*tup))) for subl in lis for tup in subl ]

输出：

[['A', '1,21', 'io'],
 ['B', '5', 'io'],
 ['A', '3', 'io'],
 ['C', '0', 'io'],
 ['D', '6', 'io'],
 ['A', '9.0', 'io']]

Answer 2

你可以尝试这个：

new_list = [[nested[0] for nested in sub_l] for l in lis for sub_l in l]
print(new_list)
[['A', '1,21', 'io'],
 ['B', '5', 'io'],
 ['A', '3', 'io'],
 ['C', '0', 'io'],
 ['D', '6', 'io'],
 ['A', '9.0', 'io']]

它不是很易读，但它可以满足您的需求。

<强>更新如果您想拥有一个字符串列表，可以使用以下代码：

[",".join(nested[0] for nested in sub_l) for l in lis for sub_l in l]

如何解压缩以下嵌套的元组列表的前值？

2 个答案: