Python Pandas:如果数据是NaN,则更改为0,否则在数据框中更改为1

Question

我有一个DataFrame:df如下:

 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

我想将NaN更改为0,否则在列中更改为1:age,url.我的代码如下,但这是错误的.

  import Pandas as pd

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

我想得到以下结果:

  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

谢谢你的帮助!

Answer 1

您可以使用wherewith fillna和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

或numpy.where用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

最快的解决方案:notnull和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

编辑:

我尝试修改你的解决方案:

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

时间:

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

使用np.where具有pd.notnull与替换丢失和有效元素0,并1分别为:

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