That does work thanks.
And in a for loop was my goal, so this works for me.
for i in np.unique(testdata[:,0]):
print(i, testdata[testdata[:,0]==i][:,2].mean())
(datetime.date(2013, 10, 7), 2.9333333333333336)
(datetime.date(2013, 10, 8), 5.633333333333334)
(datetime.date(2013, 10, 9), 4.266666666666667)
Roelf
From: [email protected]
[mailto:[email protected]] On Behalf Of Georgios Exarchakis
Sent: 09 October 2013 15:22
To: Discussion of Numerical Python
Subject: Re: [Numpy-discussion] select column based on another column value
I think you need this
>>> testdata[testdata[:,0]==today][:,2].mean()
4.266666666666667
>>> testdata[testdata[:,0]==yesterday][:,2].mean()
5.633333333333334
>>> testdata[testdata[:,0]==twodaysago][:,2].mean()
2.9333333333333336
On 10/09/2013 06:46 AM, Roelf Schreurs wrote:
Hi
I have the following array and want to calculate the average per day.
Import numpy as np
from datetime import date
today = date(today.year, today.month, 9)
yesterday = date(today.year, today.month, 8)
twodaysago = date(today.year, today.month, 7)
testdata = np.array([[today, "r", 3.2],[today, "r", 4.3],[today, "r",
5.3],[yesterday, "r", 6.3],[yesterday, "r", 9.3],[yesterday, "r",
1.3],[twodaysago, "r", 3.3],[twodaysago, "r", 1.2],[twodaysago, "r", 4.3]])
which produces
array([[datetime.date(2013, 10, 9), 'r', 3.2],
[datetime.date(2013, 10, 9), 'r', 4.3],
[datetime.date(2013, 10, 9), 'r', 5.3],
[datetime.date(2013, 10, 8), 'r', 6.3],
[datetime.date(2013, 10, 8), 'r', 9.3],
[datetime.date(2013, 10, 8), 'r', 1.3],
[datetime.date(2013, 10, 7), 'r', 3.3],
[datetime.date(2013, 10, 7), 'r', 1.2],
[datetime.date(2013, 10, 7), 'r', 4.3]], dtype=object)
And I want the output as (numbers aren't correct below):
2013-10-09, 4.54
2013-10-08, 5.43
2013-10-07, 2.76
But not sure how.
Thanks
Roelf
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