CODEWITHSUNDEEP
pythonlistpandasdictionary
rares balta
rares balta

Reputation: 19

how to output pandas in python row/column

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

The output is: 

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

But I want to be like: 

192.168.70.150    30
192.168.72.15     38
192.168.72.150    29

What do I need to change in the code?

Upvotes: 0

Views: 39

Answers (4)

A.Kot
A.Kot

Reputation: 7903

Eazy Peazy

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

Upvotes: 0

jezrael
jezrael

Reputation: 862611

If possible dupicated ip, better is create tuples in 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

Upvotes: 0

Dmitrii K
Dmitrii K

Reputation: 267

When you create df from dictionary, then the keys will be columns and the values will be rows. Use list for your goals.

>>> 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

Upvotes: 1

BENY
BENY

Reputation: 323226

You need 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

Or you can flatten your list of dict to a dict , then using 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

Upvotes: 1

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