Find rows in numpy array of matplotlib date objects

Question

I'm using matplotlib.dates to convert my string dates into date objects thinking it would be easier to manipulate later.

import matplotlib.dates as md    
def ConvertDate(datestr):
    '''
    Convert string date into matplotlib date object
    '''
    datefloat = md.datestr2num(datestr)
    return md.num2date(datefloat)

What I was trying to do was filter my structured array to tell me the index numbers of rows belong to a certain month and/or year

import numpy as np
np.where( data['date'] == 2008 )

I can probably use a lambda function to convert each object into string value like so

lambda x: x.strftime('%Y')

to compare each item but I dont know where to put this lambda function into np.where or if its even possible.

Any ideas? Or is there some better way to do this?

kentwait · Accepted Answer · 2012-12-17 11:40:18Z

1

After a lot of error messages, I think I found an answer to my own question.

[ x for x in range(len(data)) if data['date'][x].year == 2008 ]

I did a list comprehension to return the indexes of the structured array that matched a query. I also included @hayden's suggestion to use .year instead of strftime() Maybe numpy.where() is still faster but this suits my needs right now.

edited Dec 17, 2012 at 11:40

answered Dec 17, 2012 at 11:28

kentwait

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1 Comment

Andy Hayden Over a year ago

You could also use, i for i,x in enumerate(data) ... which I think reads cleaner :)

Community · Accepted Answer · 2017-05-23 11:43:06Z

1

Note: you might as well use datetime's datetime.strptime function:

import datetime
import numpy as np
dt1 = datetime.datetime.strptime('1/2/2012', '%d/%m/%Y')
dt2 = datetime.datetime.strptime('1/2/2011', '%d/%m/%Y')

In [5]: dt1
Out[5]: datetime.datetime(2012, 2, 1, 0, 0)

You can then use numpy.non-zero (to filter your array to the indices of those datetimes where, for example, year is 2012):

a = np.array([dt1, dt2])
b = np.array(map(lambda x: x.year, a))

In [8]: b
Out[8]: array([2012, 2011], dtype=bool)

In [9]: np.nonzero(b==2012)
Out[9]: (array([0]),)

Also, I would suggest looking into pandas which has this functionality built-in (on top of numpy), many more convenience functions (e.g. to_datetime), as well as efficient datetime storage...

edited May 23, 2017 at 11:43

CommunityBot

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answered Dec 17, 2012 at 11:03

Andy Hayden

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4 Comments

kentwait Over a year ago

So I think I did the opposite: I created a new array that converted the object into text newArray = np.array(map(lambda x:x.strftime('%Y'),data['date'])) and did the matching on that matches = np.where( newArray == '2008') Therefore the match indexes are also the indexes of the original array

Andy Hayden Over a year ago

My code does exactly the same thing? But I think it should be more efficient to use x.year == 2008 (since it avoid calling strftime).

kentwait Over a year ago

Sorry I'm confused with np.where(lambda x: x.year == 2012) part. Shouldn't the lambda function map to dt when you create a new array? Anyway you are right that pandas may be better for my purpose.

Andy Hayden Over a year ago

Sorry, didn't properly verify what I was doing! There must be a way without using map...

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