Python If statement on dataframe

Question

I would like to replace column df['pred'] with 0 if the respective value of df['nonzero'] is not 'NAN' and "<= 1".

         beta0  beta1  number_repair   t  pred  nonzero  
0          NaN    NaN            NaN   6     0      NaN  
1          NaN    NaN            NaN   7     0      NaN  
2          NaN    NaN            NaN   8     0      NaN  
3          NaN    NaN            NaN   9     3      0  
4          NaN    NaN            NaN  10     2      0  
5          NaN    NaN            NaN  11     1      0

I tried the following code but it returned error. How could I correct the code or could someone suggest other way to achieve it? Thanks!

mapping['pred'] = 0 if (np.all(np.isnan(mapping['nonzero'])),
(mapping['nonzero'] <= 1)) else mapping['pred']

Do you need to use a ternary? Have you tried just looping over the rows? — OneCricketeer
– OneCricketeer, Commented Apr 19, 2016 at 13:03

Community · Accepted Answer · 2017-05-23 12:31:13Z

3

I think you can use loc with mask by function notnull:

mask = (df['nonzero'].notnull()) & (df['nonzero'] <= 1)
print mask
0    False
1    False
2    False
3     True
4     True
5     True
Name: nonzero, dtype: bool

By comment (Thank you PhilChang) it is same as:

mask = df['nonzero'] <= 1
print mask
0    False
1    False
2    False
3     True
4     True
5     True
Name: nonzero, dtype: bool

df.loc[ mask, 'pred'] = 0
print df
   beta0  beta1  number_repair   t  pred  nonzero
0    NaN    NaN            NaN   6     0      NaN
1    NaN    NaN            NaN   7     0      NaN
2    NaN    NaN            NaN   8     0      NaN
3    NaN    NaN            NaN   9     0      0.0
4    NaN    NaN            NaN  10     0      0.0
5    NaN    NaN            NaN  11     0      0.0

Another solution with mask:

df['pred'] = df.pred.mask(mask,0)
print df
   beta0  beta1  number_repair   t  pred  nonzero
0    NaN    NaN            NaN   6     0      NaN
1    NaN    NaN            NaN   7     0      NaN
2    NaN    NaN            NaN   8     0      NaN
3    NaN    NaN            NaN   9     0      0.0
4    NaN    NaN            NaN  10     0      0.0
5    NaN    NaN            NaN  11     0      0.0

edited May 23, 2017 at 12:31

CommunityBot

11 silver badge

answered Apr 19, 2016 at 13:03

jezrael

868k103 gold badges1.4k silver badges1.3k bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

PhilChang Over a year ago

In fact, notnull() is not necessary in this case. float number compared with nan is always false

ysearka · Accepted Answer · 2016-04-19 13:09:17Z

0

I don't know how to check on a Series if cells contain 'NaN', but for the other condition, this works quite well:

df.ix[df.ix[:,'nonzero'] <=1,'pred'] = 0

You then just have to add after the first test "and my_second_test".

answered Apr 19, 2016 at 13:09

ysearka

3,8655 gold badges24 silver badges42 bronze badges

Collectives™ on Stack Overflow

Python If statement on dataframe

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related