Reputation: 2093
Let's say I have a pandas dataframe with two columns, a string and a datetime, like the following:
ORDER TIMESTAMP
GO 6/4/2019 09:59:49.497000
STAY 6/4/2019 09:05:27.036000
WAIT 6/4/2019 10:33:05.645000
GO 6/4/2019 10:28:03.649000
STAY 6/4/2019 11:23:11.614000
GO 6/4/2019 11:00:33.574000
WAIT 6/4/2019 11:41:55.744000
I want to create a list where each entry is a list with three values. For each time interval of choice (say one hour), each entry is: [beginning time, total number of rows, percent of rows with order GO].
For example, for the dataframe above, my list would be:
[6/4/2019 09:00:00.000000, 2, 50]
[6/4/2019 10:00:00.000000, 2, 50]
[6/4/2019 11:00:00.000000, 3, 33.3]
I created a simple while loop:
go= []
while t<=df["timestamp"].iloc[-1]:
tmp1 = df[(df["date_time"]>=t) & (df["timestamp"]<t+timedelta(hour=1))]
tmp2 = df[(df["date_time"]>=t) & (df["timestamp"]<t+timedelta(hour=1)) & (df["Order"]=="GO")]
go.append([t, tmp1.shape[0], 100.0*tmp2.shape[0]/tmp1.shape[0]])
#increment the time by the interval
t=t+timedelta(hour=1)
However, my initial dataframe has millions of rows, and I would like my time interval to be much shorter than an hour, so this approach is VERY slow. What is the more pythonic way to do it?
Upvotes: 0
Views: 183
Reputation: 150735
Let's try groupby().agg()
with size
for number of rows and mean
to get the ratio of rows with GO
:
(df.ORDER.eq('GO').astype(int)
.groupby(df.TIMESTAMP.dt.floor('1H')) # groupby interval of choice
.agg(['size','mean'])
.reset_index() # get timestamp back
.to_numpy().tolist() # this is to generate the list
)
Output:
[[Timestamp('2019-06-04 09:00:00'), 2, 0.5],
[Timestamp('2019-06-04 10:00:00'), 2, 0.5],
[Timestamp('2019-06-04 11:00:00'), 3, 0.3333333333333333]]
Upvotes: 2