将熊猫df中的不同长度长度的元素拆分字符串

发布于 2025-01-21 21:01:20 字数 1093 浏览 0 评论 0原文

我有一个看起来像这个

ID	human_id
1	（'苹果'，'2022-12-04'，'a5ted'）
2	（'Bananas'，'2012-2-14'）
3	（'2012-2-14）的数据框'，'reda21'，'ss'）
..	..

我想要一种“ Pythonic”的方法，使ID

ID ID	human_id	col1	col2 col2	col3
1	（'apples'，'2022-12-04'，'a5ted'）	苹果	2022-12-04	A5TED
2	（'Bananas'，'2012-2-14'）	香蕉	2022-12-04	NP.NAN
3	（'2012-2-14'，'reda21'，'ss'，'SS'）	2012-2 -14	reda21	：

import pandas as pd

df['a'], df['b'], df['c'] = df.human_id.str

我尝试给我错误的代码

valueerror：没有足够的值解开包装（预期2，获得1）python

如何将元组中的值拆分为列？

谢谢。

原文

I have a dataframe that looks like this

id	human_id
1	('apples', '2022-12-04', 'a5ted')
2	('bananas', '2012-2-14')
3	('2012-2-14', 'reda21', 'ss')
..	..

I would like a "pythonic" way to have such output

id	human_id	col1	col2	col3
1	('apples', '2022-12-04', 'a5ted')	apples	2022-12-04	a5ted
2	('bananas', '2012-2-14')	bananas	2022-12-04	np.NaN
3	('2012-2-14', 'reda21', 'ss')	2012-2-14	reda21	ss

import pandas as pd

df['a'], df['b'], df['c'] = df.human_id.str

The code I have tried give me error:

ValueError: not enough values to unpack (expected 2, got 1) Python

How can I split the values in tuple to be in columns?

Thank you.

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云胡 2025-01-28 21:01:20

您可以这样做。它只会在找不到值的地方放置。然后，您可以将DF1附加到DF。

d = {'id': [1,2,3], 
     'human_id': ["('apples', '2022-12-04', 'a5ted')", 
                  "('bananas', '2012-2-14')",
                  "('2012-2-14', 'reda21', 'ss')"
                 ]}

df = pd.DataFrame(data=d)

list_human_id = tuple(list(df['human_id']))

newList = []
for val in listh:
    newList.append(eval(val))

df1 = pd.DataFrame(newList, columns=['col1', 'col2', 'col3'])

print(df1)

Output


        col1        col2   col3
0     apples  2022-12-04  a5ted
1    bananas   2012-2-14   None
2  2012-2-14      reda21     ss

You can do it this way. It will just put None in places where it couldn't find the values. You can then append the df1 to df.

d = {'id': [1,2,3], 
     'human_id': ["('apples', '2022-12-04', 'a5ted')", 
                  "('bananas', '2012-2-14')",
                  "('2012-2-14', 'reda21', 'ss')"
                 ]}

df = pd.DataFrame(data=d)

list_human_id = tuple(list(df['human_id']))

newList = []
for val in listh:
    newList.append(eval(val))

df1 = pd.DataFrame(newList, columns=['col1', 'col2', 'col3'])

print(df1)

Output


        col1        col2   col3
0     apples  2022-12-04  a5ted
1    bananas   2012-2-14   None
2  2012-2-14      reda21     ss

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墨洒年华 2025-01-28 21:01:20

你可以做

out = df.join(pd.DataFrame(df.human_id.tolist(),index=df.index,columns=['a','b','c']))

You can do

out = df.join(pd.DataFrame(df.human_id.tolist(),index=df.index,columns=['a','b','c']))

回复收藏 0 原文

岁吢 2025-01-28 21:01:20

列将使用元组长度创建动态，并使用相同的dataFrame

import pandas as pd

id = [1, 2, 3]
human_id = [('apples', '2022-12-04', 'a5ted')
            ,('bananas', '2012-2-14')
            , ('2012-2-14', 'reda21', 'ss')]

df = pd.DataFrame({'id': id, 'human_id': human_id})

print("*"*20,'Dataframe',"*"*20)
print(df.to_string())

print()

print("*"*20,'Split Data',"*"*20)

row = 0

for x in df['human_id']:

    col = 1

    for xx in x:

        #df['col'+str(z)] = str(xx)

        name_column = 'col'+str(col)
        df.loc[df.index[row], name_column] = str(xx)

        col+=1

    row+=1

print(df.to_string())

column will create dynamic with length of tuple and using same dataframe

import pandas as pd

id = [1, 2, 3]
human_id = [('apples', '2022-12-04', 'a5ted')
            ,('bananas', '2012-2-14')
            , ('2012-2-14', 'reda21', 'ss')]

df = pd.DataFrame({'id': id, 'human_id': human_id})

print("*"*20,'Dataframe',"*"*20)
print(df.to_string())

print()

print("*"*20,'Split Data',"*"*20)

row = 0

for x in df['human_id']:

    col = 1

    for xx in x:

        #df['col'+str(z)] = str(xx)

        name_column = 'col'+str(col)
        df.loc[df.index[row], name_column] = str(xx)

        col+=1

    row+=1

print(df.to_string())