通过替换空值python将嵌套列表合并为一个

Question

我有一个列表，里面有几个嵌套列表，这里是示例：

temp_lst = [[None, None, None, None, None, None, None, '51', '51', '19'],
            [None, None, None, '55', '55', '55', '55', None, None, None],
            ['23', '23', '55', None, None, None, None, None, None, None]]

我想把它合并到一个列表中，并且值应该按照特定的顺序，结果需要完全像这样：

           ['23', '23', '55', '55', '55', '55', '55', '51', '51', '19']

temp_lst 中可能有更多列表。

Answer 1

你需要转置你的矩阵，然后过滤掉可能的候选者。

arr=[[None, None, None, None, None, None, None, '51', '51', '19'],
 [ None, None, None,'55', '55', '55', '55', None, None, None],
['23', '23', '55',None, None, None,None, None, None,None]]

ans=[]
for i in zip(*arr[::-1]):
    ans.append(list(filter(lambda x:x is not None, i))[0])
print(ans)

Answer 2

试试这个：

new_list = [n for sublist in original_list for n in sublist if n != None]

基本上我在这里所做的是使用列表推导。

Answer 3

您并没有真正扁平化您的列表，而是从子列表中的每个位置获取不是None 的值。

您可以使用zip() 按位置组合您的列表，并使用or 来获得不是None（或'' - 你没有）的值：

# you keep the vaue on each position thats not None 
data =  [[None, None, None, None, None, None, None, '51', '51', '19'],
         [None, None, None, '55', '55', '55', '55', None, None, None],
         ['23', '23', '55', None, None, None, None, None, None, None]]
# wanted ['23', '23', '55', '55', '55', '55', '55', '51', '51', '19']

def coalesque(iterable, default):
    """Returns the first truthy value from iterable or default."""
    for x in iterable:
        if x: 
            return x
    return default

result = [ coalesque(n,'') for n in zip(*data)]  

print(result)

将输出：

['23', '23', '55', '55', '55', '55', '55', '51', '51', '19']

只要您能确保：

• 所有列表的长度都相同（其他研究 itertools.zip_longest）
• 每个“列”只有 1 个不是 None 的元素（仅使用第一个）

---

使用default 和coalesque(iterable,default) 来定义如果一个“列”中的所有值都是虚假的，应该返回什么。

见：

• truth value testing
• itertools.zip_longest 用于大小不一的列表

Answer 4

您可以使用 numpy 来执行此操作。 使用 Numpy，您可以展平任何维度的列表

import numpy as np
a  = [[None, None, None, None, None, None, None, '51', '51', '19'],
 [ None, None, None,'55', '55', '55', '55', None, None, None],
['23', '23', '55',None, None, None,None, None, None,None]]
a = np.array(a) # Converting to numpy array
a = a.flatten() # Flatten a list with numpy's builtin function
a = [i for i in a if i != None] # Removing all occurences of None
a = list(a) # Converting back to python list

通过，您可以展平任何维度的列表

Answer 5

虽然其他人已经发布了一个衬里，但如果您的列表中有更多层，这里有一个递归函数将起作用。

def unstack_list(temp_list):
    # Construct list of elements to return
    answer = []
    for i in temp_list:
        # if element of list is list, recurse and concatenate lists
        if type(i) == list:
            answer = unstack_list(i) + answer
        # add to list if not None
        elif i is not None:
            answer.append(i)
    return answer

Answer 6

此解决方案保留每行的最后一个非 None 值，并跳过显示所有 None 值的任何位置。

def crunchnones(data):
"""
Assumes data is an iterable containing lists of values, and
    returns a single list of all non-None value in all lists in data,
    maintaining their positions inside each list.
If no list contain a non-None value in a given position, that position
    will be silently skipped.
If more than one list contains a non-None value in a given position,
    only the one in the last list will be returned.

Example:
>>> data =  [[None, None, None, None, None, None, None, '51', '51', '19'],
             [None, None, None, '55', '55', '55', '55', None, None, None],
             ['23', '23', '55', None, None, None, None, None, None, None]]
>>> data2 =  [[None, None, None, None, None, None, None, '51', None, '19'],
              [None, None, None, '55', '55', '55', '55', None, None, None],
              [None, '23', '55', None, None, None, None, None, None, None]]
>>> data3 =  [['21', None, None, None, None, None, None, '51', '51', '19'],
              ['22', None, None, '55', '55', '55', '55', None, None, None],
              ['23', '23', '55', None, None, None, None, None, None, None]]

>>>crunchnones(data)
['23', '23', '55', '55', '55', '55', '55', '51', '51', '19']
>>>crunchnones(data2)
['23', '55', '55', '55', '55', '55', '51', '19']
>>>crunchnones(data3)
['23', '23', '55', '55', '55', '55', '55', '51', '51', '19']
>>>
"""
#
temp = {}
for riga in data:
    temp.update({i: x for i, x in enumerate(riga) if x is not None})
return [temp[i] for i in sorted(temp)]