在Python中将一个字符串列拆分为多个列

Question

I have a following dataframe:

df = pd.DataFrame({'scene':[{"living":"0.515","kitchen":"0.297"}, {"kitchen":"0.401","study":"0.005"}, {"study":"0.913"}, {}, {"others":"0"}], 'id':[1, 2, 3 ,4, 5]}) 

id        scene
01      {"living":"0.515","kitchen":"0.297"}
02      {"kitchen":"0.401","study":"0.005"}
03      {"study":"0.913"}
04      {}
05      {"others":"0"}

and I want to create a new dataframe as shown below, can someone help me to create this using Pandas?

id      living     kitchen     study     others
01      0.515       0.297        0         0 
02        0         0.401      0.005       0
03        0           0        0.913       0
04        0           0          0         0 
05        0           0          0         0

Answer 1

Simple solution is to convert your scene column to the list of dictionaries and create new data frame with default constructor:

pd.DataFrame(df.scene.tolist()).fillna(0)

Result:

  kitchen living others  study
0   0.297  0.515      0      0
1   0.401      0      0  0.005
2       0      0      0  0.913
3       0      0      0      0
4       0      0      0      0

One of the "default" way to create DataFrame is to use a list of dictionaries. In this case each dictionary of list will be converted to the separate row and each key of dict will be used for the column heading.

Answer 2

On your data,

df = pd.DataFrame({'scene':[{"living":"0.515","kitchen":"0.297"}, {"kitchen":"0.401","study":"0.005"}, 
                        {"study":"0.913"}, {}, {"others":"0"}], 
               'id':[1, 2, 3 ,4,5], 's': ['a','b','c','d','e']})

df:
    id  s   scene
0   1   a   {'kitchen': '0.297', 'living': '0.515'}
1   2   b   {'kitchen': '0.401', 'study': '0.005'}
2   3   c   {'study': '0.913'}
3   4   d   {}
4   5   e   {'others': '0'}

There are two ways you can go about doing this,

1. In a single line, where you have to input all column names except 'scene' to set_index function

    df = df.set_index(['id', 's'])['scene'].apply(pd.Series).fillna(0).reset_index() 

   which will output:

     id s kitchen living study others 0 1 a 0.297 0.515 0 0 1 2 b 0.401 0 0.005 0 2 3 c 0 0 0.913 0 3 4 d 0 0 0 0 4 5 e 0 0 0 0 

2. In two lines, where you create your excepted result and concat it to the original dataframe.

    df1 = df.scene.apply(pd.Series).fillna(0) df = pd.concat([df, df1], axis=1) 

   which gives,

     id s scene kitchen living study others 0 1 a {'kitchen': '0.297', 'living': '0.515'} 0.297 0.515 0 0 1 2 b {'kitchen': '0.401', 'study': '0.005'} 0.401 0 0.005 0 2 3 c {'study': '0.913'} 0 0 0.913 0 3 4 d {} 0 0 0 0 4 5 e {'others': '0'} 0 0 0 0 

Answer 3

Updated. This one works perfectly. Welcome to give your suggestions to keep it more concise.

import json
import pandas as pd

df = pd.DataFrame({'scene':[{"living":"0.515","kitchen":"0.297"}, {"kitchen":"0.401","study":"0.005"}, {"study":"0.913"}, {}, {"others":"0"}], 'id':[1, 2, 3 ,4,5], 's':['a','b','c','d','e']}) 
def test(Scene, type):
    Scene = json.loads(Scene)
    if type in Scene.keys():
        return Scene[type]
    else:
        return ""

a = ['living', 'kitchen', 'study', 'others']
for b in a:
    df[b] = df['Scene'].map(lambda Scene: test(Scene, b.lower()))

cols = ['living', 'kitchen', 'study', 'others']
df[cols] = df[cols].replace({'': 0})
df[cols] = df[cols].apply(pd.to_numeric, errors='coerce', axis=1)

Answer 4

完美的一线解决方案就在这里，感谢所有帮助：

df.join(df['scene'].apply(json.loads).apply(pd.Series))