Python 向具有可变长度的数据帧添加新列

关于将结果添加到现有数据帧，我有一个问题

if relevant_item != 'None' and relevant_item != 'Not in dict':
    items = relevant_item
    len_item = len(items)

    if len_item == 1:
        item_result = items

    if len_item == 2:
        two = items
        item_result = some_method(two)

    if len_item == 3:
        threes = items
        item_result = some_method(three)

hash_in_dict_shopping.append(item_result)#new list of list

shops = pd.Series(hash_in_dict_shopping)
df_final['hash_in_shop'] = shops.values

在将新列表添加到现有数据帧时，我收到一条错误消息“ValueError:值的长度与索引的长度不匹配”，因此我想知道如何将新列表添加到新列，并在保持原始顺序的同时用“无”填充所有缺少的行值

过滤前的原始数据（约700行）：

过滤相关项目的数据后（大约40行）：

应用某些_方法后（从dict返回值）：

数据框中包含所有700行的新列：

'None'
'None'
['fruit','green groceries']
'None'
'None'
'None'
['dry food', 'staples', 'legumes']
'None'
'None'
['dairy']

---

有两点需要注意：

迭代序列时，不应忽略/跳过“无”/“不在目录中”行。新系列的长度必须与原始系列的长度相同

您应该使用内置功能来按行应用函数。由于您不能使用矢量化功能（因为您的数据框包含

列表

对象），因此您可以使用带有自定义功能的

pd.Series.apply

下面是一个简单的例子：

df = pd.DataFrame({'col': ['None', 'Not in dict', ['apple', 'banana', 'grapes'],
                           'None', ['mile'], 'Not in dict']})

def calculated(x):
    try:
        if x in {'Not in dict', 'None'}:
            return None
    except TypeError:
        if len(x) == 1:
            return 2
        elif len(x) == 2:
            return 4
        else:
            return 6

df['calc'] = df['col'].apply(calculated)

print(df)

                       col  calc
0                     None   NaN
1              Not in dict   NaN
2  [apple, banana, grapes]   6.0
3                     None   NaN
4                   [mile]   2.0
5              Not in dict   NaN

---

您是否尝试设置空数组，然后更改值（如果有的话）

import numpy as np
items = numpy.empty((len(DataFrame))
items[:] = numpy.nan

if relevant_item != 'None' and relevant_item != 'Not in dict':
items[i] = relevant_item # supposing you have some so
len_item = count_nonzero(np.isnan(items))

if len_item == 1:
    item_result = items

if len_item == 2:
    two = items
    item_result = some_method(two)

if len_item == 3:
    threes = items
    item_result = some_method(three)

这样，您的items数组的长度与数据帧的长度相同，您将不会得到该错误。如果NaN数组不合适，为什么不尝试numpy.zeros

希望这有帮助

'None'
'None'
['fruit','green groceries']
'None'
'None'
'None'
['dry food', 'staples', 'legumes']
'None'
'None'
['dairy']

df = pd.DataFrame({'col': ['None', 'Not in dict', ['apple', 'banana', 'grapes'],
                           'None', ['mile'], 'Not in dict']})

def calculated(x):
    try:
        if x in {'Not in dict', 'None'}:
            return None
    except TypeError:
        if len(x) == 1:
            return 2
        elif len(x) == 2:
            return 4
        else:
            return 6

df['calc'] = df['col'].apply(calculated)

print(df)

                       col  calc
0                     None   NaN
1              Not in dict   NaN
2  [apple, banana, grapes]   6.0
3                     None   NaN
4                   [mile]   2.0
5              Not in dict   NaN

import numpy as np
items = numpy.empty((len(DataFrame))
items[:] = numpy.nan

if relevant_item != 'None' and relevant_item != 'Not in dict':
items[i] = relevant_item # supposing you have some so
len_item = count_nonzero(np.isnan(items))

if len_item == 1:
    item_result = items

if len_item == 2:
    two = items
    item_result = some_method(two)

if len_item == 3:
    threes = items
    item_result = some_method(three)