Python 如何使用pandas中现有列之一的列表中的名称创建新列，并从另一列的列表中指定值？

我对大数据帧有一个问题，大约1kk行，180列。它以3列开始。第一列包含ID。第二行和第三行包含每行中的列表-它们是连接的（第一行-第一列列表中的第一个元素与第二列列表中的第一个元素连接：

ids | fruits | count |

1 | [grape, apple, banana]  | [7.0, 4.0, 3.0]

2 | [mango, banana, strawberry, grape] | [5.0, 8.0, 15.0, 2.0]

3 | [apple, avocado] | [9.0, 1.0]
4 | NaN | NaN
5 | [pummelo] | [12.0]

我想使用'fruits'列中的列表元素作为新列的名称，这些新列的值将分配给row和fruits。但列不能重复，例如：

ids | grape | apple | banana | mango | strawberry | avocado | pummelo

1 | 7.0 | 4.0 | 3.0 | NaN | NaN | NaN | NaN

2 | 2.0 | NaN | 8.0 | 5.0 | 15.0 | NaN | NaN

3 | NaN | 9.0 | NaN | NaN | NaN | 1.0 | NaN

4 | NaN | NaN | NaN | NaN | NaN | NaN | NaN

5 | NaN | NaN | NaN | NaN | NaN | NaN | 12.0

集合中唯一元素的计数（所有列表的非重复和）'fruits'是180，这就是为什么我希望最后有180列

问题是速度。我尝试了pandas iterrows（），但当涉及到所有1kk行时，它就变成了无休止的故事。下面是我尝试过的代码

#making an example dataframe

import numpy as np
fruit_df = pd. DataFrame(columns=['ids','fruits','count'])
ids = [1,2,3,4,5]
fruits = [['grape', 'apple', 'banana'], ['mango', 'banana', 'strawberry', 'grape'], ['apple', 'avocado'], np.nan, ['pummelo']]
count = [[7.0, 4.0, 3.0],[5.0, 8.0, 15.0, 2.0], [9.0, 1.0], np.nan, [12.0]]


#creating fruits columns in dataframe - this one timing is ok , fine for me (about 15 mins)

fruits_columns=[]
for row in fruit_df['fruits']:
    if type(row)==list:
        fruits_columns.append(row)
    else:
        fruits_columns.append(list())

import itertools
all_fruits = list(itertools.chain(*fruits_columns))

all_fruits = set(all_fruits)

for fruit in all_fruits:
    fruit_df[fruit]=np.nan


#iterating over the data - here is main problem - takes very, very long time.. works well for this tiny dataset but when it comes to 1000000 rows and 180 columns...

def iter_over_rows(data):
    for index, row in data.iterrows():
        if type(row['fruits'])!=float:
            for cat in range(len(row['fruits'])):       
                data[row['fruits'][cat]][index] = row['count'][cat]

---

我想加快数据处理的速度。考虑过用所有180个水果作为键并将它们作为值来制作字典，但在最后，顺序会被破坏。如果你知道如何更快地处理，那就太好了。干杯！

这会做你想做的一切，但它会删除

ID4

，因为它们只会同时使用ntain

NA

值

设置相同：

fruit_df = pd. DataFrame(columns=['ids','fruits','count'])
ids = [1,2,3,4,5]
fruits = [['grape', 'apple', 'banana'], ['mango', 'banana', 'strawberry', 'grape'], ['apple', 'avocado'], np.nan, ['\
pummelo']]
count = [[7.0, 4.0, 3.0],[5.0, 8.0, 15.0, 2.0], [9.0, 1.0], np.nan, [12.0]]

fruit_df['ids'] = ids
fruit_df['fruits'] = fruits
fruit_df['count'] = count

我们希望将包含列表的行转换为堆叠系列（这基本上只是将列表扩展为新行，同时保留行的ID:

fruit_df.set_index(['ids'], inplace=True)
fruit_series = fruit_df.apply(lambda x: pd.Series(x['fruits']), axis=1).stack()
count_series = fruit_df.apply(lambda x: pd.Series(x['count']), axis=1).stack()

final_df = pd.DataFrame()
final_df['Fruits'] = fruit_series
final_df['Counts'] = count_series
print(final_df)

所以我们看到最终的_df看起来像：

           Fruits  Counts
ids
1   0       grape     7.0
    1       apple     4.0
    2      banana     3.0
2   0       mango     5.0
    1      banana     8.0
    2  strawberry    15.0
    3       grape     2.0
3   0       apple     9.0
    1     avocado     1.0
5   0     pummelo    12.0

好的，很酷，现在我们已经扩展了列表行以匹配它们的id，但是我们现在看到了这个我们不想要的multi_索引df，所以我们将删除它，然后旋转我们的表，使id成为索引，并生成列：

final_df = final_df.reset_index().drop('level_1', axis=1)
final_df = final_df.pivot(index='ids', columns = 'Fruits', values = 'Counts')
print(final_df)

返回：

Fruits  apple avocado banana grape mango pummelo strawberry
ids
1         4.0     NaN    3.0   7.0   NaN     NaN        NaN
2         NaN     NaN    8.0   2.0   5.0     NaN       15.0
3         9.0     1.0    NaN   NaN   NaN     NaN        NaN
5         NaN     NaN    NaN   NaN   NaN    12.0        NaN

非常接近，我希望这对你有用！ 整个代码组合：

import pandas as pd
import numpy as np

fruit_df = pd. DataFrame(columns=['ids','fruits','count'])
ids = [1,2,3,4,5]
fruits = [['grape', 'apple', 'banana'], ['mango', 'banana', 'strawberry', 'grape'], ['apple', 'avocado'], np.nan, ['\
pummelo']]
count = [[7.0, 4.0, 3.0],[5.0, 8.0, 15.0, 2.0], [9.0, 1.0], np.nan, [12.0]]

fruit_df['ids'] = ids
fruit_df['fruits'] = fruits
fruit_df['count'] = count


fruit_df.set_index(['ids'], inplace=True)
fruit_series = fruit_df.apply(lambda x: pd.Series(x['fruits']), axis=1).stack()
count_series = fruit_df.apply(lambda x: pd.Series(x['count']), axis=1).stack()

final_df = pd.DataFrame()

final_df['Fruits'] = fruit_series
final_df['Counts'] = count_series

final_df = final_df.reset_index().drop('level_1', axis=1)
final_df = final_df.pivot(index='ids', columns = 'Fruits', values = 'Counts')

print(final_df)

---

这将完成您想要的所有操作，但它将删除IDs4，因为它们只包含NA值

设置相同：

fruit_df = pd. DataFrame(columns=['ids','fruits','count'])
ids = [1,2,3,4,5]
fruits = [['grape', 'apple', 'banana'], ['mango', 'banana', 'strawberry', 'grape'], ['apple', 'avocado'], np.nan, ['\
pummelo']]
count = [[7.0, 4.0, 3.0],[5.0, 8.0, 15.0, 2.0], [9.0, 1.0], np.nan, [12.0]]

fruit_df['ids'] = ids
fruit_df['fruits'] = fruits
fruit_df['count'] = count

我们希望将包含列表的行转换为堆叠系列（这基本上只是将列表扩展为新行，同时保留行的ID:

fruit_df.set_index(['ids'], inplace=True)
fruit_series = fruit_df.apply(lambda x: pd.Series(x['fruits']), axis=1).stack()
count_series = fruit_df.apply(lambda x: pd.Series(x['count']), axis=1).stack()

final_df = pd.DataFrame()
final_df['Fruits'] = fruit_series
final_df['Counts'] = count_series
print(final_df)

所以我们看到最终的_df看起来像：

           Fruits  Counts
ids
1   0       grape     7.0
    1       apple     4.0
    2      banana     3.0
2   0       mango     5.0
    1      banana     8.0
    2  strawberry    15.0
    3       grape     2.0
3   0       apple     9.0
    1     avocado     1.0
5   0     pummelo    12.0

好的，很酷，现在我们已经扩展了列表行以匹配它们的id，但是我们现在看到了这个我们不想要的multi_索引df，所以我们将删除它，然后旋转我们的表，使id成为索引，并生成列：

final_df = final_df.reset_index().drop('level_1', axis=1)
final_df = final_df.pivot(index='ids', columns = 'Fruits', values = 'Counts')
print(final_df)

返回：

Fruits  apple avocado banana grape mango pummelo strawberry
ids
1         4.0     NaN    3.0   7.0   NaN     NaN        NaN
2         NaN     NaN    8.0   2.0   5.0     NaN       15.0
3         9.0     1.0    NaN   NaN   NaN     NaN        NaN
5         NaN     NaN    NaN   NaN   NaN    12.0        NaN

非常接近，我希望这对你有用！ 整个代码组合：

import pandas as pd
import numpy as np

fruit_df = pd. DataFrame(columns=['ids','fruits','count'])
ids = [1,2,3,4,5]
fruits = [['grape', 'apple', 'banana'], ['mango', 'banana', 'strawberry', 'grape'], ['apple', 'avocado'], np.nan, ['\
pummelo']]
count = [[7.0, 4.0, 3.0],[5.0, 8.0, 15.0, 2.0], [9.0, 1.0], np.nan, [12.0]]

fruit_df['ids'] = ids
fruit_df['fruits'] = fruits
fruit_df['count'] = count


fruit_df.set_index(['ids'], inplace=True)
fruit_series = fruit_df.apply(lambda x: pd.Series(x['fruits']), axis=1).stack()
count_series = fruit_df.apply(lambda x: pd.Series(x['count']), axis=1).stack()

final_df = pd.DataFrame()

final_df['Fruits'] = fruit_series
final_df['Counts'] = count_series

final_df = final_df.reset_index().drop('level_1', axis=1)
final_df = final_df.pivot(index='ids', columns = 'Fruits', values = 'Counts')

print(final_df)

---

工作得很好！我在设置之后用替换的方式处理了NaNs：fruit_df['fruits']=fruit_df['fruits'].应用（lambda d:d如果是instance（d，list）或者其他['no_fruit']）fruit_df['count']=fruit_df['count'].应用（lambda d:d如果是instance（d，list）或者[-1]），然后再次替换它们（对我来说，重要的是只保留NaNs的数据）：final_df=final_df.replace（to_replace=['not_four']，value=np.nan）final_df=final_df.replace（to_replace=-1，value=np.nan）谢谢你的回答，我给了+1，但似乎我在这里太新了，它被计算了，但没有显示这是处理NaN's的一个好方法！我很高兴你发现这个答案很有用，并且我很高兴处理这个问题。效果很好！我在设置后用替换的方式处理NaN's：fruit_df['fruits']=fruit_df['fruits']。申请（lambda d:d如果isinstance（d，list）else['no_fruit']）fruit_df['count']=fruit_df['count']）应用（lambda d:d如果isinstance（d，list）else[-1]），然后再次替换它们（对我来说，重要的是仅用nans保存数据）：final_df=final_df.replace（to_replace=['not_fruit']，value=np.nan）final_df=final_df.replace（to_replace=-1，value=np.nan）谢谢你的回答，我给出了+1，但似乎我在这里太新了，它被计算了，但没有显示这是处理nan的一个好方法！我很高兴你发现答案很有用，并且我在这方面很开心。