熊猫数据框任意两列之间的百分比差异

Question

I would like to have a function defined for percentage diff calculation between any two pandas columns. Lets say that my dataframe is defined by:

R1  R2    R3    R4   R5    R6
 A   B     1     2    3     4

I would like my calculation defined as

df['R7'] = df[['R3','R4']].apply( method call to calculate perc diff)

and

df['R8'] = df[['R5','R6']].apply(same method call to calculate perc diff)

How to do?

I have tried below

df['perc_cnco_error'] = df[['CumNetChargeOffs_x','CumNetChargeOffs_y']].apply(lambda x,y: percCalc(x,y))

def percCalc(x,y):
    if x<1e-9:
        return 0
    else:
        return (y - x)*100/x

and it gives me the error message

TypeError: ('() takes exactly 2 arguments (1 given)', u'occurred at index CumNetChargeOffs_x')

Answer 1

At it's simplest terms, is this what you're looking for?

def percentage_change(col1,col2):
    return ((col2 - col1) / col1) * 100

You can apply it to any 2 columns of your dataframe:

df['a'] = percentage_change(df['R3'],df['R4'])    
df['b'] =  percentage_change(df['R6'],df['R5'])

Out[220]: 
  R1 R2  R3  R4  R5  R6      a     b
0  A  B   1   2   3   4  100.0 -25.0

Answer 2

This would give you the deviation in percentage:

df.apply(lambda row: (row.iloc[0]-row.iloc[1])/row.iloc[0]*100, axis=1)

If you have more than two columns try,

df[['R3', 'R5']].apply(lambda row: (row.iloc[0]-row.iloc[1])/row.iloc[0]*100, axis=1)

Answer 3

要计算R3和R4之间的百分比差异，您可以使用：

df['R7'] = (df.R3 - df.R4) / df.R3 * 100