Lambda including if…elif…else

Question or problem about Python programming:

I want to apply a lambda function to a DataFrame column using if…elif…else within the lambda function.

The df and the code are smth. like:

df=pd.DataFrame({"one":[1,2,3,4,5],"two":[6,7,8,9,10]})

df["one"].apply(lambda x: x*10 if x<2 elif x<4 x**2 else x+10)

obviously this way it is not working.
Is there a way to apply if....elif....else to lambda?
How can I relize the same result with List Comprehension?

Thanks for any response.

How to solve the problem:

Solution 1:

Nest if .. elses:

lambda x: x*10 if x<2 else (x**2 if x<4 else x+10)

Solution 2:

I do not recommend the use of apply here: it should be avoided if there are better alternatives.

For example, if you are performing the following operation on a Series:

if cond1:
    exp1
elif cond2:
    exp2
else:
    exp3

This is usually a good use case for np.where or np.select.

---

numpy.where

The if else chain above can be written using

np.where(cond1, exp1, np.where(cond2, exp2, ...))

np.where allows nesting. With one level of nesting, your problem can be solved with,

df['three'] = (
    np.where(
        df['one'] < 2, 
        df['one'] * 10, 
        np.where(df['one'] < 4, df['one'] ** 2, df['one'] + 10))
df

   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

---

numpy.select

Allows for flexible syntax and is easily extensible. It follows the form,

np.select([cond1, cond2, ...], [exp1, exp2, ...])

Or, in this case,

np.select([cond1, cond2], [exp1, exp2], default=exp3)

df['three'] = (
    np.select(
        condlist=[df['one'] < 2, df['one'] < 4], 
        choicelist=[df['one'] * 10, df['one'] ** 2], 
        default=df['one'] + 10))
df

   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

---

and/or (similar to the if/else)

Similar to if-else, requires the lambda:

df['three'] = df["one"].apply(
    lambda x: (x < 2 and x * 10) or (x < 4 and x ** 2) or x + 10) 

df
   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

---

List Comprehension

Loopy solution that is still faster than apply.

df['three'] = [x*10 if x<2 else (x**2 if x<4 else x+10) for x in df['one']]
# df['three'] = [
#    (x < 2 and x * 10) or (x < 4 and x ** 2) or x + 10) for x in df['one']
# ]
df
   one  two  three
0    1    6     10
1    2    7      4
2    3    8      9
3    4    9     14
4    5   10     15

Solution 3:

For readability I prefer to write a function, especially if you are dealing with many conditions. For the original question:

def parse_values(x):
    if x < 2:
       return x * 10
    elif x < 4:
       return x ** 2
    else:
       return x + 10

df['one'].apply(parse_values)

Hope this helps!