Python的大熊猫日期时间差

Question

我有这样的：

x[l[0]] = pd.to_datetime(x[l[0]], format="%Y-%m-%d %H:%M:%S") 

凡l=list(x)

我怎么能在几秒钟内这之间的区别对象，如果我这样做

x[l[0]][1]-x[l[0]][2] 

它返回我timedelta对象

print (x[:5]) 
       LogDate Query_BoxID_ID Query_Function_ID SC_Win32_Status 
0 2017-06-15 09:50:14    12     24    0 
1 2017-06-15 09:50:14    12     26    0 
2 2017-06-15 09:50:14    12     26    0 
3 2017-06-15 09:50:14    12     30    0 
4 2017-06-15 09:50:32    12     19    0 

Answer 1

我可以做

(x[l[0]][1]-x[l[0]][2]).total_seconds() 

Answer 2

使用diff为timedeltas，这是由total_seconds转换：

#convert column to datetime 
x['LogDate'] = pd.to_datetime(x['LogDate'], format="%Y-%m-%d %H:%M:%S") 
#first value is NaN always, so replaced to 0 by fillna and cast to int 
a = x['LogDate'].diff().dt.total_seconds().fillna(0).astype(int) 
print (a) 
0  0 
1  0 
2  0 
3  0 
4 18 
Name: LogDate, dtype: int32 

---

b = int((x.loc[0, 'LogDate'] - x.loc[0, 'LogDate']).total_seconds()) 
print (b) 
0