[英]how to speed up this calculation?
i have a data set that looks like this:我有一个看起来像这样的数据集:
|.....userId.................|..cahtroomID....|..msg_index_in_chat..|..time_difference_between_msg..| |...userId.....|..cahtroomID..|..msg_index_in_chat..|..time_difference_between_msg..| |1234567891222222|sdfbsjkfdsdklf...|..............1.................|......0 hours 0 minutes....................|
|1234567891222222|sdfbsjkfdsdklf...|................1................................|......0 小时 0分钟………………| |9876543112252141|sdfbsjkfdsdklf...|...............2................|......0 hours 4 minutes....................|
|9876543112252141|sdfbsjkfdsdklf...|................2................|......0 小时 4分钟………………| |2374623982398939|quweioqewiieio|...............1................|......0 hours 0 minutes....................|
|2374623982398939|quweioqewiieio|................1................|......0 小时 0 分钟.. ...................| |1234567891222222|quweioqewiieio|...............2................|......0 hours 4 minutes....................|
|1234567891222222|quweioqewiieio|................2................|......0 小时 4 分钟.. ...................| |2374623982398939|quweioqewiieio|....................3...........|......1 hours 0 minutes....................|
|2374623982398939|quweioqewiieio|..................3..........|......1 小时 0 分钟.. ...................|
I need to calculate the average time between messages in every room and assign the value I've gotten to every row.我需要计算每个房间中消息之间的平均时间,并将我得到的值分配给每一行。 To do so, I wrote this:
为此,我写了这个:
df['avg_time'] = 0
for room in set(df.roomId):
table = df[['msg_index_in_chat', 'time_difference_between_msg']][df.roomId == room]
if len(table) > 1:
avg_time = []
times = table.time_difference_between_msg.tolist()
avg_time = sum(times[1:], timedelta(0))/len(times[1:])
elif len(table) ==1:
avg_time = timedelta(hours = 0)
df.loc[df.roomId == room,('avg_time')] = avg_time
the problem is that this code runs for a lot of time.问题是这段代码运行了很长时间。 can you suggest a faster way for doing this calculation?
你能建议一个更快的方法来做这个计算吗?
Thank you!谢谢!
Use GroupBy.transform
with custom lambda function:将
GroupBy.transform
与自定义 lambda 函数一起使用:
f = lambda times: sum(times.iloc[1:], pd.Timedelta(0))/len(times.iloc[1:]) if len(times) > 1 else pd.Timedelta(0)
df['avg_time'] = df.groupby('cahtroomID')['time_difference_between_msg'].transform(f)
print (df)
userId cahtroomID msg_index_in_chat \
0 1234567891222222 sdfbsjkfdsdklf 1
1 9876543112252141 sdfbsjkfdsdklf 2
2 2374623982398939 quweioqewiieio 1
3 1234567891222222 quweioqewiieio 2
4 2374623982398939 quweioqewiieio 3
time_difference_between_msg avg_time
0 00:00:00 00:04:00
1 00:04:00 00:04:00
2 00:00:00 00:32:00
3 00:04:00 00:32:00
4 01:00:00 00:32:00
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