将不同长度的嵌套列表的 Python 字典导出到 csv。如果嵌套列表有 > 1 个条目，在移动到下一个键之前扩展到列

Question

我有以下列表字典

d = {1: ['1','B1',['C1','C2','C3']], 2: ['2','B2','C15','D12'], 3: ['3','B3'], 4: ['4', 'B4', 'C4', ['D1', 'D2']]}

使用

将其写入 csv

with open('test.csv', "w", newline = '') as f:
writer = csv.writer(f)
writer.writerow(headers)
writer.writerows(d.values())

给我一个看起来像

的 csv

A     B            C                D
1    B1     ['C1','C2',C3']
2    B2           C15             D12
3    B3
4    B4            C4          ['D1','D2']

如果值中有多个项目列表（嵌套列表？），我希望该列表像这样向下展开列

A     B     C      D
1    B1    C1
1          C2 
1          C3
2    B2    C15    D12
3    B3
4    B4    C4      D1
4                  D2

我是 python 的新手，经过几天的论坛筛选和苦苦思索后，我似乎无法找到一种方法来做我需要做的事情。我想我可能需要拆分嵌套列表，但我需要让它们与各自的 "A" 值相关联。 A 列和 B 列始终有 1 个条目，C 列和 D 列可以有 1 到 X 个条目。

非常感谢任何帮助

Answer 1

似乎制作一个列表的列表可能比您正在做的更容易，并在适当的位置留出空格。这是可能做的事情：

import csv
from itertools import zip_longest

def condense(dct):
    # get the maximum number of columns of any list
    num_cols = len(max(dct.values(), key=len)) - 1
    # Ignore the key, it's not really relevant.
    for _, v in dct.items():
        # first, memorize the index of this list, 
        # since we need to repeat it no matter what
        idx = v[0]
        # next, use zip_longest to make a correspondence.
        # We will deliberately make a 2d list, 
        # and we will later withdraw elements from it one by one.
        matrix = [([]     if elem is None else
                   [elem] if not isinstance(elem, list) else 
                   elem[:]  # soft copy to avoid altering original dict
                  ) for elem, _ in zip_longest(v[1:], range(num_cols), fillvalue=None)
                 ]
        # Now, we output the top row of the matrix as long as it has contents
        while any(matrix):
            # If a column in the matrix is empty, we put an empty string.
            # Otherwise, we remove the row as we pass through it,
            # progressively emptying the matrix top-to-bottom
            # as we output a row, we also remove that row from the matrix.
            # *-notation is more convenient than concatenating these two lists.
            yield [idx, *((col.pop(0) if col else '') for col in matrix)] 
            # e.g. for key 0 and a matrix that looks like this:
            #    [['a1', 'a2'],
            #     ['b1'],
            #     ['c1', 'c2', 'c3']]
            # this would yield the following three lists before moving on:
            #    ['0', 'a1', 'b1', 'c1']
            #    ['0', 'a2', '',   'c2']
            #    ['0', '',   '',   'c3']
            # where '' should parse into an empty column in the resulting CSV.

这里要注意的最重要的事情是我使用 isinstance(elem, list) 作为 shorthand 来检查这个东西是否是一个列表（你需要能够以某种方式做到这一点，像我们在这里做的那样压扁或圆化列表）。如果您有更复杂或更多样的数据结构，则需要通过此检查即兴发挥 - 也许编写一个辅助函数 isiterable() 尝试迭代并 returns 一个布尔值，基于这样做是否产生一个错误。

完成后，我们可以在 d 上调用 condense() 并让 csv 模块处理输出。

headers = ['A', 'B', 'C', 'D']
d = {1: ['1','B1',['C1','C2','C3']], 2: ['2','B2','C15','D12'], 3: ['3','B3'], 4: ['4', 'B4', 'C4', ['D1', 'D2']]}

# condense(d) produces
#   [['1', 'B1', 'C1',  ''   ], 
#    ['1', '',   'C2',  ''   ], 
#    ['1', '',   'C3',  ''   ], 
#    ['2', 'B2', 'C15', 'D12'], 
#    ['3', 'B3', '',     ''  ], 
#    ['4', 'B4', 'C4',  'D1' ], 
#    ['4', '',   '',    'D2' ]]

with open('test.csv', "w", newline = '') as f:
    writer = csv.writer(f)
    writer.writerow(headers)
    writer.writerows(condense(d))

生成以下文件：

A,B,C,D
1,B1,C1,
1,,C2,
1,,C3,
2,B2,C15,D12
3,B3,,
4,B4,C4,D1
4,,,D2

这相当于您的预期输出。希望该解决方案具有足够的可扩展性，以便您将其应用于您的非 MVCE 问题。

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