在大熊猫删除列不需要的内容（而不是整个列）

Question

我应该怎么做，如果我只是想在timestamp列显示2016-06-29拥有并删除不需要的部分像15:46:43.895000

Answer 1

如果时间戳：

df['timestamp'] = df['timestamp'].dt.floor('d') 

如果字符串：

df['timestamp'] = df['timestamp'].str.split().str[0] 

样品：

df = pd.DataFrame({'timestamp':pd.date_range('2016-06-29 15:46:43.895000', 
           periods=3, 
           freq='2000T')}) 
print (df) 
       timestamp 
0 2016-06-29 15:46:43.895 
1 2016-07-01 01:06:43.895 
2 2016-07-02 10:26:43.895 

print (type(df.loc[0, 'timestamp'])) 
<class 'pandas._libs.tslib.Timestamp'> 

df['timestamp'] = df['timestamp'].dt.floor('d') 
print (df) 
    timestamp 
0 2016-06-29 
1 2016-07-01 
2 2016-07-02 

---

df = pd.DataFrame({'timestamp':['2016-06-20 15:46:43.895000', 
           '2016-06-22 15:46:43.895000', 
           '2016-06-29 15:46:43.895000']}) 
print (df) 
        timestamp 
0 2016-06-20 15:46:43.895000 
1 2016-06-22 15:46:43.895000 
2 2016-06-29 15:46:43.895000 

print (type(df.loc[0, 'timestamp'])) 
<class 'str'> 

df['timestamp'] = df['timestamp'].str.split().str[0] 
print (df) 
    timestamp 
0 2016-06-20 
1 2016-06-22 
2 2016-06-29 

Answer 2

我们可以使用.dt.normalize()方法：

df['timestamp'] = df['timestamp'].dt.normalize() 

Answer 3

df['timestamp']=pd.to_datetime(df['timestamp']).dt.strftime('%Y-%m-%d')