Could Pandas Use Column as Index

Could pandas use column as index?

Yes, with pandas.DataFrame.set_index you can make 'Locality' your row index.

data.set_index('Locality', inplace=True)

If inplace=True is not provided, set_index returns the modified dataframe as a result.

Example:

> import pandas as pd
> df = pd.DataFrame([['ABBOTSFORD', 427000, 448000],
                     ['ABERFELDIE', 534000, 600000]],
                    columns=['Locality', 2005, 2006])

> df
     Locality    2005    2006
0  ABBOTSFORD  427000  448000
1  ABERFELDIE  534000  600000

> df.set_index('Locality', inplace=True)
> df
              2005    2006
Locality                  
ABBOTSFORD  427000  448000
ABERFELDIE  534000  600000

> df.loc['ABBOTSFORD']
2005    427000
2006    448000
Name: ABBOTSFORD, dtype: int64

> df.loc['ABBOTSFORD'][2005]
427000

> df.loc['ABBOTSFORD'].values
array([427000, 448000])

> df.loc['ABBOTSFORD'].tolist()
[427000, 448000]

Pandas index column title or name

You can just get/set the index via its name property

In [7]: df.index.name
Out[7]: 'Index Title'

In [8]: df.index.name = 'foo'

In [9]: df.index.name
Out[9]: 'foo'

In [10]: df
Out[10]: 
         Column 1
foo              
Apples          1
Oranges         2
Puppies         3
Ducks           4

In pandas, how to set_index with using column index instead of referring to column names?

If the column index is unique you could use:

df.set_index(list(df.columns[cols]))

where cols is a list of ordinal indices.

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For example,

In [77]: np.random.seed(2016)

In [79]: df = pd.DataFrame(np.random.randint(10, size=(5,4)), columns=list('ABCD'))

In [80]: df
Out[80]: 
   A  B  C  D
0  3  7  2  3
1  8  4  8  7
2  9  2  6  3
3  4  1  9  1
4  2  2  8  9

In [81]: df.set_index(list(df.columns[[0,2]]))
Out[81]: 
     B  D
A C      
3 2  7  3
8 8  4  7
9 6  2  3
4 9  1  1
2 8  2  9

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If the DataFrame's column index is not unique, then setting the index by label
is impossible and by ordinals more complicated:

import numpy as np
import pandas as pd
np.random.seed(2016)

def set_ordinal_index(df, cols):
    columns, df.columns = df.columns, np.arange(len(df.columns))
    mask = df.columns.isin(cols)
    df = df.set_index(cols)
    df.columns = columns[~mask]
    df.index.names = columns[mask]
    return df

df = pd.DataFrame(np.random.randint(10, size=(5,4)), columns=list('AAAA'))
print(set_ordinal_index(df, [0,2]))

yields

     A  A
A A      
3 2  7  3
8 8  4  7
9 6  2  3
4 9  1  1
2 8  2  9

Get column index from column name in python pandas

Sure, you can use .get_loc():

In [45]: df = DataFrame({"pear": [1,2,3], "apple": [2,3,4], "orange": [3,4,5]})

In [46]: df.columns
Out[46]: Index([apple, orange, pear], dtype=object)

In [47]: df.columns.get_loc("pear")
Out[47]: 2

although to be honest I don't often need this myself. Usually access by name does what I want it to (df["pear"], df[["apple", "orange"]], or maybe df.columns.isin(["orange", "pear"])), although I can definitely see cases where you'd want the index number.

python - How to set column as index in a DataFrame

I think you need transpose by T and then if necessary change column names add list comprehension:

Notice:
Double groupby+sum is not necessary, once is enough, because aggregate same aggregate function, here sum.

df = pd.DataFrame(salary_List)

newdf = df.groupby(pd.Grouper(key='date', freq='1M')).sum().T
#python 3.6+
newdf.columns = [f'Month{x}' for x in range(1, len(newdf.columns) + 1)]

#python bellow
#newdf.columns = ['Month{}'.format(x) for x in range(1, len(newdf.columns) + 1)]
print (newdf)
                        Month1   Month2    Month3    Month4    Month5  \
Balance before Salary     27.2     88.2    176.48     48.48    241.48   
Salary                 15300.0  15300.0  14783.00  16249.00  14448.00   

                         Month6  
Balance before Salary     49.48  
Salary                 15663.00  

Pandas (python): How to add column to dataframe for index?

How about this:

from pandas import *

idx = Int64Index([171, 174, 173])
df = DataFrame(index = idx, data =([1,2,3]))
print df

It gives me:

     0
171  1
174  2
173  3

Is this what you are looking for?

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