Question

import pandas as pd

tabel = [{'192.168.70.150': '30'}, 
         {'192.168.72.15': '38'}, 
         {'192.168.72.150': '29'}]
df = pd.DataFrame(tabel)
print df

输出结果为：

     192.168.70.150 192.168.72.15 192.168.72.150             
      0             30           NaN            NaN                
      1            NaN            38            NaN               
      2            NaN           NaN             29

但我希望如此：

192.168.70.150    30
192.168.72.15     38
192.168.72.150    29

我需要在代码中更改什么？

Answer 1

您需要stack

df.stack()
Out[349]: 
0  192.168.70.150    30
1  192.168.72.15     38
2  192.168.72.150    29
dtype: object

或者你可以将你的dict列表展平为dict，然后使用pd.Serise

df = pd.Series({k: v for d in tabel for k, v in d.items()})
df
Out[353]: 
192.168.70.150    30
192.168.72.15     38
192.168.72.150    29
dtype: object

Answer 2

从字典创建df时，键将是列，值将是行。使用列表来实现目标。

>>> import pandas as pd
>>> tabel = [['192.168.70.150', '30'], 
...          ['192.168.72.15', '38'], 
...          ['192.168.72.150', '29']]
>>> df = pd.DataFrame(table, columns=['IP', 'Value'])
>>> df
               IP Value
0  192.168.70.150 30
1  192.168.72.15  38
2  192.168.72.150 29

Answer 3

如果可能重复ip，最好在tuple中创建list comprehension：

tabel = [{'192.168.70.150': '30'}, 
             {'192.168.72.15': '38'}, 
             {'192.168.72.150': '29'},
             {'192.168.72.150': '20'}]

L = [(a, b) for d in tabel for a, b in d.items()]
df = pd.DataFrame(L, columns=['a','b'])
print (df)
                a   b
0  192.168.70.150  30
1   192.168.72.15  38
2  192.168.72.150  29
3  192.168.72.150  20

Answer 4

Eazy Peazy

pd.DataFrame([list(i.items())[0] for i in tabel])

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