Python Pandas: if the data is NaN, then change to be 0, else change to be 1 in data frame

Question

I have a DataFrame:df as following:

 row  id  name    age   url           
  1   e1   tom    NaN   http1   
  2   e2   john   25    NaN
  3   e3   lucy   NaN  http3 
  4   e4   tick   29    NaN

I want to change the NaN to be 0, else to be 1 in the columns: age, url. My code is following, but it is wrong.

  import Pandas as pd

  df[['age', 'url']].applymap(lambda x: 0 if x=='NaN' else x)

I want to get the following result:

  row  id  name    age   url           
  1   e1   tom     0     1
  2   e2   john    1     0
  3   e3   lucy    0     1 
  4   e4   tick    1     0

Thanks for your help!

Answer 1

You can use where with fillna and condition by isnull:

df[['age', 'url']] = df[['age', 'url']].where(df[['age', 'url']].isnull(), 1)
                                       .fillna(0).astype(int)
print (df)

   row  id  name  age  url
0    1  e1   tom    0    1
1    2  e2  john    1    0
2    3  e3  lucy    0    1
3    4  e4  tick    1    0

Or numpy.where with isnull:

df[['age', 'url']] = np.where(df[['age', 'url']].isnull(), 0, 1)
print (df)
   row  id  name  age  url
0    1  e1   tom    0    1
1    2  e2  john    1    0
2    3  e3  lucy    0    1
3    4  e4  tick    1    0

Fastest solution with notnull and astype:

df[['age', 'url']] = df[['age', 'url']].notnull().astype(int)
print (df)
   row  id  name  age  url
0    1  e1   tom    0    1
1    2  e2  john    1    0
2    3  e3  lucy    0    1
3    4  e4  tick    1    0

EDIT:

I try modify your solution:

df[['age', 'url']] = df[['age', 'url']].applymap(lambda x: 0 if pd.isnull(x) else 1)
print (df)
   row  id  name  age  url
0    1  e1   tom    0    1
1    2  e2  john    1    0
2    3  e3  lucy    0    1
3    4  e4  tick    1    0

Timings:

len(df)=4k:

In [127]: %timeit df[['age', 'url']] = df[['age', 'url']].applymap(lambda x: 0 if pd.isnull(x) else 1)
100 loops, best of 3: 11.2 ms per loop

In [128]: %timeit df[['age', 'url']] = np.where(df[['age', 'url']].isnull(), 0, 1)
100 loops, best of 3: 2.69 ms per loop

In [129]: %timeit df[['age', 'url']] = np.where(pd.notnull(df[['age', 'url']]), 1, 0)
100 loops, best of 3: 2.78 ms per loop

In [131]: %timeit df.loc[:, ['age', 'url']] = df[['age', 'url']].notnull() * 1
1000 loops, best of 3: 1.45 ms per loop

In [136]: %timeit df[['age', 'url']] = df[['age', 'url']].notnull().astype(int)
1000 loops, best of 3: 1.01 ms per loop

Answer 2

Use np.where with pd.notnull to replace the missing and valid elements with 0 and 1 respectively:

In [90]:
df[['age', 'url']] = np.where(pd.notnull(df[['age', 'url']]), 1, 0)
df

Out[90]:
   row  id  name  age  url
0    1  e1   tom    0    1
1    2  e2  john    1    0
2    3  e3  lucy    0    1
3    4  e4  tick    1    0

Answer 3

df.loc[:, ['age', 'url']] = df[['age', 'url']].notnull() * 1
df