handing exceptions based on a particular data frame column

Question

I have a data frame as follows:

Now I want to divide the x by y values without giving me an exception. For example when I divide 3 by 2 it should give me 1.5 and when I divide 3 by 0 it should give me zero. To achieve this I have written an exception function

def divide(x,y):
    try:
        result = x/y
        print (result)
    except ZeroDivisionError:
        print (0)

I want to now create a new column in the data-frame and apply this function. So far I have used:

df['num']  = df.apply(divide)

But this is not giving me the required result. Can someone help

use return x/y instead of print

kofemann
– kofemann

2019-04-17 06:32:15 +00:00
Commented Apr 17, 2019 at 6:32 — kofemann
– kofemann, Commented Apr 17, 2019 at 6:32

Arundathi · Accepted Answer · 2019-04-17 07:05:41Z

2

You can also try:

def divide(df):
    if(df['y']!=0):
        return df['x']/df['y']
    else:
        return 0
df['num']=df.apply(divide,axis=1)

But I would suggest using this instead:

import numpy as np
df['num']=df['x']/df['y']
df.replace(np.inf,0)

answered Apr 17, 2019 at 7:05

Arundathi

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Comments

Umesh · Accepted Answer · 2019-04-17 10:50:11Z

1

Try this:

if you are using python 2.7 then return x/float(y)

import math

def divide(x,y):
        # if you check type(y) or type(x) you will get type: numpy.float64
        # pandas internally converts it to a numpy.float64
        # when you do division on such values you will get inf as output
        # you can check if values are not zero and do calculations or convert it to float like: x / float(y)

        if y and not math.isnan(y):
            return x/y

df['num']  = df.apply(lambda row: divide(row["x"], row["y"]), axis=1)

edited Apr 17, 2019 at 10:50

answered Apr 17, 2019 at 6:38

Umesh

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7 Comments

Umesh Over a year ago

more on apply method: pandas.pydata.org/pandas-docs/stable/reference/api/…

Omer Qureshi Over a year ago

I'm using python 3.7

Omer Qureshi Over a year ago

I ran the code but it's giving me inf values when a number is being divided by 0. Perhaps the divide function is not working properly

Omer Qureshi Over a year ago

x and y are both present in my df. the formula works when dividing numbers like 17 by 3 but then it divides 11 by 0 it gives me inf values. To counter this we wrote the exception formula. I wonder why that is not working

Umesh Over a year ago

inf is not a ZeroDivisionError. it's just giving an answer as infinite., Check data type of x and y. Is it numpy.float64 or silimar rather than float/int?

|

U13-Forward · Accepted Answer · 2019-04-17 06:38:51Z

0

You need to return the stuff, not print the stuff, so it doesn't just displays, but also stores in memory:

def divide(x,y):
    try:
        return x/y
    except ZeroDivisionError:
        return 0

df['num'] = df.apply(lambda x: divide(x["x"], x["y"]), axis=1)

edited Apr 17, 2019 at 6:38

answered Apr 17, 2019 at 6:33

U13-Forward

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3 Comments

Omer Qureshi Over a year ago

Okay. Thanks for pointing out my mistake. Now if I want to create a new column and apply it on my df it is still giving me an error. I'm using : df['num'] = df.apply(divide)

U13-Forward Over a year ago

@OmerQureshi edited my answer, please accept and up-vote it if it works.

Omer Qureshi Over a year ago

I ran the code but it is giving me inf for numbers which are dividing by 0. This is exactly what I wanted to avoid.

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