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python pandas operations on columns


python pandas operations on columns

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python pandas operations on columns
Tag : python , By : silvervino
Date : November 25 2020, 07:22 PM


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Python pandas data frame: how to perform operations on two columns with the same name


Tag : python , By : Carlos Galdino
Date : March 29 2020, 07:55 AM
Hope this helps You can use iloc if you dont want rename columns:
import numpy as np
import pandas as pd

np.random.seed(0)
df = pd.DataFrame(np.random.rand(4,5), columns = list('abcab'))
print df
          a         b         c         a         b
0  0.548814  0.715189  0.602763  0.544883  0.423655
1  0.645894  0.437587  0.891773  0.963663  0.383442
2  0.791725  0.528895  0.568045  0.925597  0.071036
3  0.087129  0.020218  0.832620  0.778157  0.870012
#select first a column
print df.iloc[:,0]
0    0.548814
1    0.645894
2    0.791725
3    0.087129
Name: a, dtype: float64

#select second a column
print df.iloc[:,3]
Name: a, dtype: float64
0    0.544883
1    0.963663
2    0.925597
3    0.778157
Name: a, dtype: float64

#select first a column
print df['a'].iloc[:,0]
0    0.548814
1    0.645894
2    0.791725
3    0.087129
Name: a, dtype: float64

#select second a column
print df['a'].iloc[:,1]
0    0.544883
1    0.963663
2    0.925597
3    0.778157
Name: a, dtype: float64
import numpy as np
import pandas as pd

np.random.seed(0)
df = pd.DataFrame(np.random.rand(4,5), columns = list('abbab'))
print df
          a         b         b         a         b
0  0.548814  0.715189  0.602763  0.544883  0.423655
1  0.645894  0.437587  0.891773  0.963663  0.383442
2  0.791725  0.528895  0.568045  0.925597  0.071036
3  0.087129  0.020218  0.832620  0.778157  0.870012

cols=pd.Series(df.columns)
for dup in df.columns.get_duplicates():
    cols[df.columns.get_loc(dup)]=[dup+'_'+str(d_idx) if d_idx!=0 else dup for d_idx in range(df.columns.get_loc(dup).sum())]
df.columns=cols
print df
          a         b       b_1       a_1       b_2
0  0.548814  0.715189  0.602763  0.544883  0.423655
1  0.645894  0.437587  0.891773  0.963663  0.383442
2  0.791725  0.528895  0.568045  0.925597  0.071036
3  0.087129  0.020218  0.832620  0.778157  0.870012

Pandas operations on selected columns


Tag : python , By : HokieGeek
Date : March 29 2020, 07:55 AM
should help you out I want to standardize certain columns in my pandas dataframe. , You could concatenate the scaled columns to the original DF as shown:
scaler = StandardScaler()
scaled_data = pd.DataFrame(data=scaler.fit_transform(dfTest[['A', 'B']]), 
                           columns=['A_scaled', 'B_scaled'])

pd.concat([dfTest, scaled_data], axis=1)

Pandas Columns Operations with List


Tag : python , By : rusl
Date : March 29 2020, 07:55 AM
wish of those help Your problem was that you were always updating the whole column with the value of the last calculation with these lines:
df['over_360'] = len(over_360)
df['under_360'] = len(under_360)
df.set_value(i,'over_360',len(over_360))
df.set_value(i,'under_360',len(under_360))
df.ix[i,'over_360'] = len(over_360)
df.ix[i,'under_360'] = len(under_360)

operations with columns in pandas


Tag : python , By : Murali Ravipudi
Date : March 29 2020, 07:55 AM
To fix this issue Use filter for select columns starts with I:
df1 = df.filter(regex='^I')
print (df1)
   I_0  I_1  I_2
q               
0    2    3    4
1    5    4    3
2    4    8    7
df2 = df[["10", "20", "30"]].mul(df["SF"], axis=0)
print (df2)
    10   20   30
q               
0  1.0  0.2  0.2
1  1.2  2.1  2.4
2  3.6  3.2  2.8
df = df1.sub(df2.values, 0).add_suffix('_sub')
print (df)
   I_0_sub  I_1_sub  I_2_sub
q                           
0      1.0      2.8      3.8
1      3.8      1.9      0.6
2      0.4      4.8      4.2
print (df1.sub(df2, axis=0))
   10  20  30  I_0  I_1  I_2
q                           
0 NaN NaN NaN  NaN  NaN  NaN
0 NaN NaN NaN  NaN  NaN  NaN
2 NaN NaN NaN  NaN  NaN  NaN

Pandas Select Columns with simple operations


Tag : python , By : Barry
Date : March 29 2020, 07:55 AM
Hope this helps You need to use .assign()
C['final'] = C.assign(num1_aux=C['num1']/2)[['num1_aux','num2']].min(1)
    name    num1    num2    final
0   A        99      3.5    3.5
1   B         3     99.0    1.5
2   C         5     -5.0    -5.0
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