if else function in pandas dataframe [duplicate]

Answer 1

There is another way without np.where or pd.Series.where. Am not saying it is better, but after trying to adapt this solution to a challenging problem today, was finding the syntax for where no so intuitive. In the end, not sure whether it would have possible with where, but found the following method lets you have a look at the subset before you modify it and it for me led more quickly to a solution. Works for the OP here of course as well.

You deliberately set a value on a slice of a dataframe as Pandas so often warns you not to.

This answer shows you the correct method to do that.

The following gives you a slice:

df.loc[df['age1'] - df['age2'] > 0]

..which looks like:

   age1  age2
0    23    10
1    45    20

Add an extra column to the original dataframe for the values you want to remain after modifying the slice:

df['diff'] = 0

Now modify the slice:

df.loc[df['age1'] - df['age2'] > 0, 'diff'] = df['age1'] - df['age2']

..and the result:

   age1  age2  diff
0    23    10    13
1    45    20    25
2    21    50     0

Answer 2

You can use numpy.where:

def my_fun (var1,var2,var3):
    df[var3]= np.where((df[var1]-df[var2])>0, df[var1]-df[var2], 0)
    return df

df1 = my_fun('age1','age2','diff')
print (df1)
   age1  age2  diff
0    23    10    13
1    45    20    25
2    21    50     0

Error is better explain here.

Slowier solution with apply, where need axis=1 for data processing by rows:

def my_fun(x, var1, var2, var3):
    print (x)
    if (x[var1]-x[var2])>0 :
        x[var3]=x[var1]-x[var2]
    else:
        x[var3]=0
    return x    

print (df.apply(lambda x: my_fun(x, 'age1', 'age2','diff'), axis=1))
   age1  age2  diff
0    23    10    13
1    45    20    25
2    21    50     0

Also is possible use loc, but sometimes data can be overwritten:

def my_fun(x, var1, var2, var3):
    print (x)
    mask = (x[var1]-x[var2])>0
    x.loc[mask, var3] = x[var1]-x[var2]
    x.loc[~mask, var3] = 0

    return x    

print (my_fun(df, 'age1', 'age2','diff'))
   age1  age2  diff
0    23    10  13.0
1    45    20  25.0
2    21    50   0.0

Answer 3

You can use pandas.Series.where

df.assign(age3=(df.age1 - df.age2).where(df.age1 > df.age2, 0))

   age1  age2  age3
0    23    10    13
1    45    20    25
2    21    50     0

---

You can wrap this in a function

def my_fun(v1, v2):
    return v1.sub(v2).where(v1 > v2, 0)

df.assign(age3=my_fun(df.age1, df.age2))

   age1  age2  age3
0    23    10    13
1    45    20    25
2    21    50     0