将内部0和第一个索引附加到第二个索引的二维列表中的所有元素- python

Question

你好，这里是python的新手……想知道解决这样的问题最好的方法是什么。

我有一个二维数组，看起来像这样：

a = [['October 17', 'Manhattan', '10024, 10025, 10026'], 
     ['October 17', 'Queen', '11360, 11362, 11365, 11368']]

我想通过迭代来创建一个新的列表或数据框，如下所示：

10024, October 17, Manhattan
10025, October 17, Manhattan
10026, October 17, Manhattan
11360, October 17, Queens
11362, October 17, Queens
11365, October 17, Queens
11368, October 17, Queens

任何见解都将不胜感激。

谢谢。

Answer 1

您可能需要迭代这些值，对于每个值，还需要迭代您拥有的几个索引

values = [['October 17', 'Manhattan', '10024, 10025, 10026'],
          ['October 17', 'Queens', '11360, 11362, 11365, 11368']]

result = [[int(idx), row[0], row[1]]
          for row in values
          for idx in row[2].split(',')]
df = DataFrame(result, columns=['idx', 'date', 'place'])

要获得

[[10024, 'October 17', 'Manhattan'], [10025, 'October 17', 'Manhattan'], 
 [10026, 'October 17', 'Manhattan'], [11360, 'October 17', 'Queens'], 
 [11362, 'October 17', 'Queens'], [11365, 'October 17', 'Queens'], 
 [11368, 'October 17', 'Queens']]


     idx        date      place
0  10024  October 17  Manhattan
1  10025  October 17  Manhattan
2  10026  October 17  Manhattan
3  11360  October 17     Queens
4  11362  October 17     Queens
5  11365  October 17     Queens
6  11368  October 17     Queens

Answer 2

Azro的答案很好，但这里有一种更明确的方法来解决这个问题(如果您是Python的新手，这种方法可能会更清楚，这样您就可以理解发生了什么-它不依赖于嵌套列表理解)。

a = [['October 17', 'Manhattan', '10024, 10025, 10026'], 
['October 17', 'Queen', '11360, 11362, 11365, 11368']]

# initialize an empty list to put our reformatted data into
final_result = []

for inner_list in a:
    # get the first two items
    first_item = inner_list[0]
    second_item = inner_list[1]

    # split the third list item, and remove trailing and leading whitespace
    remaining_data = [x.strip() for x in inner_list[2].split(',')]

    # iterate over the remaining data
    for item in remaining_data:
        # first, create a new sublist containing our first two items
        new_entry = [item, first_item, second_item]

        # add our new_entry into the final result list
        final_result.append(new_entry)

print(final_result)