Python: Maintain multiple columns with np.where

Question

Is it possible to maintain multiple columns at once using np.where? usually, one column is maintained with np.where, so my coding would look like this:

df['col1'] = np.where(df[df.condition == 'yes'],'sth', '')
df['col2'] = np.where(df[df.condition == 'yes'], 50.00, 0.0)

But caused by the fact that I test for the same condition twice, I was wondering, if I can pass 2 columns and fill them both in one run.

I tried this:

df['col1','col2'] = np.where(df[df.condition == 'yes'],['sth',50.00], ['',0.0])

But it does not work. Is there a way to realize this?

What do you mean by "maintain multiple columns at once"? Can you show a sample of the data you have and the result you are trying to achieve? — javidcf
– javidcf, Commented Mar 23, 2018 at 10:15
Possible duplicate of Numpy where function multiple conditions — Rafael
– Rafael, Commented Mar 23, 2018 at 10:15
@jezrael, I'm struggling to understand this question. If you understand it, would mind editing the question? I can see this confusing many users, not just me! — jpp
– jpp, Commented Mar 24, 2018 at 21:05

jezrael · Accepted Answer · 2018-03-25 13:15:30Z

2

I think need reshape boolean mask to (N x 1):

m = df.condition == 'yes'
df[['col1','col2']] = pd.DataFrame(np.where(m[:, None], ['sth',50.00], ['',0.0]))

Only disadvatage of solution is if different types of values in lists - numeric with strings - then numpy.where both output columns convert to strings.

Sample:

df = pd.DataFrame({'A':list('abcdef'),
                     'condition':['yes'] * 3 + ['no'] * 3})

print (df)
   A condition
0  a       yes
1  b       yes
2  c       yes
3  d        no
4  e        no
5  f        no

m = df.condition == 'yes'
df[['col1','col2']] = pd.DataFrame(np.where(m[:, None], ['sth',50.00], ['',0.0]))
print (df)
   A condition col1  col2
0  a       yes  sth  50.0
1  b       yes  sth  50.0
2  c       yes  sth  50.0
3  d        no        0.0
4  e        no        0.0
5  f        no        0.0

print (df.applymap(type))
               A      condition           col1           col2
0  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>
1  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>
2  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>
3  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>
4  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>
5  <class 'str'>  <class 'str'>  <class 'str'>  <class 'str'>

EDIT: I test it with NaNs values:

df = pd.DataFrame({'A':list('abcdefghi'),
                     'condition':['yes'] * 3 + ['no'] * 3 + [np.nan] * 3})

m = df.condition == 'yes'
df[['col1','col2']] = pd.DataFrame(np.where(m[:, None], ['sth',50.00], ['',0.0]))
print (df)
   A condition col1  col2
0  a       yes  sth  50.0
1  b       yes  sth  50.0
2  c       yes  sth  50.0
3  d        no        0.0
4  e        no        0.0
5  f        no        0.0
6  g       NaN        0.0
7  h       NaN        0.0
8  i       NaN        0.0

edited Mar 25, 2018 at 13:15

answered Mar 24, 2018 at 20:25

jezrael

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6 Comments

MaMo Over a year ago

thank you, this almost works. Any idea, why it does not work for the last 6 rows? col1 and col2 both contain nan within these last rows, all other rows have valid content.

jezrael Over a year ago

@MaMo - hmmm, it seems there is problem need another condition for exclude NaNs.

jezrael Over a year ago

@MaMo - I add sample to answer with NaNs and it working nice. Can you test it? Or I dont understand what you need?

MaMo Over a year ago

I get NaNs in col1 and col2 for last few rows, not in the condition column. I so not know the reason because there is nothing special about those rows or the condition column.

jezrael Over a year ago

@MaMo - Without data hard to know, are data confidental?

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