Python Pandas：如果数据为NaN，则更改为0，否则在数据框中更改为1
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Python Pandas: if the data is NaN, then change to be 0, else change to be 1 in data frame
提问by tktktk0711
I have a DataFrame:df as following:
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
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.
我想在列中将 NaN 更改为 0，否则更改为 1：age、url。我的代码如下，但它是错误的。
import Pandas as pd
df[['age', 'url']].applymap(lambda x: 0 if x=='NaN' else x)
I want to get the following result:
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
Thanks for your help!
谢谢你的帮助！
采纳答案by jezrael
You can use wherewith fillnaand condition by isnull:
您可以使用wherewithfillna和条件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.wherewith 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 notnulland 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:
编辑：
I try modify your solution:
我尝试修改您的解决方案：
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
Timings:
时间：
len(df)=4k:
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
回答by EdChum
Use np.wherewith pd.notnullto replace the missing and valid elements with 0and 1respectively:
使用np.wherewith分别pd.notnull用0and替换缺失的和有效的元素1：
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
回答by piRSquared
df.loc[:, ['age', 'url']] = df[['age', 'url']].notnull() * 1
df
