Numpy array conditional matching

Question

I need to match two very large Numpy arrays (one is 20000 rows, another about 100000 rows) and I am trying to build a script to do it efficiently. Simple looping over the arrays is incredibly slow, can someone suggest a better way? Here is what I am trying to do: array datesSecondDict and array pwfs2Dates contain datetime values, I need to take each datetime value from array pwfs2Dates (smaller array) and see if there is a datetime value like that (plus minus 5 minutes) in array datesSecondDict (there might be more than 1). If there is one (or more) I populate a new array (of the same size as array pwfs2Dates) with the value (one of the values) from array valsSecondDict (which is just the array with the corresponding numerical values to datesSecondDict). Here is a solution by @unutbu and @joaquin that worked for me (thanks guys!):

import time
import datetime as dt
import numpy as np

def combineArs(dict1, dict2):
   """Combine data from 2 dictionaries into a list.
   dict1 contains primary data (e.g. seeing parameter).
   The function compares each timestamp in dict1 to dict2
   to see if there is a matching timestamp record(s)
   in dict2 (plus/minus 5 minutes).
   ==If yes: a list called data gets appended with the
   corresponding parameter value from dict2.
   (Note that if there are more than 1 record matching,
   the first occuring value gets appended to the list).
   ==If no: a list called data gets appended with 0."""
   # Specify the keys to use    
   pwfs2Key = 'pwfs2:dc:seeing'
   dimmKey = 'ws:seeFwhm'

   # Create an iterator for primary dict 
   datesPrimDictIter = iter(dict1[pwfs2Key]['datetimes'])

   # Take the first timestamp value in primary dict
   nextDatePrimDict = next(datesPrimDictIter)

   # Split the second dictionary into lists
   datesSecondDict = dict2[dimmKey]['datetime']
   valsSecondDict  = dict2[dimmKey]['values']

   # Define time window
   fiveMins = dt.timedelta(minutes = 5)
   data = []
   #st = time.time()
   for i, nextDateSecondDict in enumerate(datesSecondDict):
       try:
           while nextDatePrimDict < nextDateSecondDict - fiveMins:
               # If there is no match: append zero and move on
               data.append(0)
               nextDatePrimDict = next(datesPrimDictIter)
           while nextDatePrimDict < nextDateSecondDict + fiveMins:
               # If there is a match: append the value of second dict
               data.append(valsSecondDict[i])
               nextDatePrimDict = next(datesPrimDictIter)
       except StopIteration:
           break
   data = np.array(data)   
   #st = time.time() - st    
   return data

Thanks, Aina.

11 revs · Accepted Answer · 2011-12-20 21:14:47Z

Are the array dates sorted ?

If yes, you can speed up your comparisons by breaking from the inner loop comparison once its dates are bigger than the date given by the outer loop. In this way you will made a one-pass comparison instead of looping dimVals items len(pwfs2Vals) times
If no, maybe you should transform the current pwfs2Dates array to, for example, an array of pairs [(date, array_index),...] and then you can sort by date all your arrays to make the one-pass comparison indicated above and at the same time to be able to get the original indexes needed to set data[i]

for example if the arrays were already sorted (I use lists here, not sure you need arrays for that): (Edited: now using and iterator not to loop pwfs2Dates from the beginning on each step):

pdates = iter(enumerate(pwfs2Dates))
i, datei = pdates.next() 

for datej, valuej in zip(dimmDates, dimvals):
    while datei < datej - fiveMinutes:
        i, datei = pdates.next()
    while datei < datej + fiveMinutes:
        data[i] = valuej
        i, datei = pdates.next()

Otherwise, if they were not ordered and you created the sorted, indexed lists like this:

pwfs2Dates = sorted([(date, idx) for idx, date in enumerate(pwfs2Dates)])
dimmDates = sorted([(date, idx) for idx, date in enumerate(dimmDates)])

the code would be:
(Edited: now using and iterator not to loop pwfs2Dates from the beginning on each step):

pdates = iter(pwfs2Dates)
datei, i = pdates.next()

for datej, j in dimmDates:
    while datei < datej - fiveMinutes:
        datei, i = pdates.next()
    while datei < datej + fiveMinutes:
        data[i] = dimVals[j]
        datei, i = pdates.next()

great!

..

Note that dimVals:
```
dimVals  = np.array(dict1[dimmKey]['values'])
```
is not used in your code and can be eliminated.
Note that your code gets greatly simplified by looping through the array itself instead of using xrange

Edit: The answer from unutbu address some weak parts in the code above. I indicate them here for completness:

Use of next: next(iterator) is prefered to iterator.next(). iterator.next() is an exception to a conventional naming rule that has been fixed in py3k renaming this method as iterator.__next__().
Check for the end of the iterator with a try/except. After all the items in the iterator are finished the next call to next() produces an StopIteration Exception. Use try/except to kindly break out of the loop when that happens. For the specific case of the OP question this is not an issue, because the two arrrays are the same size so the for loop finishes at the same time than the iterator. So no exception is risen. However, there could be cases were dict1 and dict2 are not the same size. And in this case there is the posibility of an exception being risen. Question is: what is better, to use try/except or to prepare the arrays before looping by equalizing them to the shorter one.

16 revs, 2 users 95% · Accepted Answer · 2017-05-23 12:20:13Z

Building on joaquin's idea:

import datetime as dt
import itertools

def combineArs(dict1, dict2, delta = dt.timedelta(minutes = 5)):
    marks = dict1['datetime']
    values = dict1['values']
    pdates = iter(dict2['datetime'])

    data = []
    datei = next(pdates)
    for datej, val in itertools.izip(marks, values):
        try:
            while datei < datej - delta:
                data.append(0)
                datei = next(pdates)
            while datei < datej + delta:
                data.append(val)
                datei = next(pdates)
        except StopIteration:
            break
    return data

dict1 = { 'ws:seeFwhm':
          {'datetime': [dt.datetime(2011, 12, 19, 12, 0, 0),
                        dt.datetime(2011, 12, 19, 12, 1, 0),
                        dt.datetime(2011, 12, 19, 12, 20, 0),
                        dt.datetime(2011, 12, 19, 12, 22, 0),
                        dt.datetime(2011, 12, 19, 12, 40, 0), ],
           'values': [1, 2, 3, 4, 5] } }
dict2 = { 'pwfs2:dc:seeing':
          {'datetime': [dt.datetime(2011, 12, 19, 12, 9),
                         dt.datetime(2011, 12, 19, 12, 19),
                         dt.datetime(2011, 12, 19, 12, 29),
                         dt.datetime(2011, 12, 19, 12, 39),
                        ], } }

if __name__ == '__main__':
    dimmKey = 'ws:seeFwhm'
    pwfs2Key = 'pwfs2:dc:seeing'    
    print(combineArs(dict1[dimmKey], dict2[pwfs2Key]))

yields

[0, 3, 0, 5]

Benjamin · Accepted Answer · 2011-12-19 19:00:40Z

I think you can do it with one fewer loops:

import datetime
import numpy

# Test data

# Create an array of dates spaced at 1 minute intervals
m = range(1, 21)
n = datetime.datetime.now()
a = numpy.array([n + datetime.timedelta(minutes=i) for i in m])

# A smaller array with three of those dates
m = [5, 10, 15]
b = numpy.array([n + datetime.timedelta(minutes=i) for i in m])

# End of test data

def date_range(date_array, single_date, delta):
    plus = single_date + datetime.timedelta(minutes=delta)
    minus = single_date - datetime.timedelta(minutes=delta)
    return date_array[(date_array < plus) * (date_array > minus)]

dates = []
for i in b:
    dates.append(date_range(a, i, 5))

all_matches = numpy.unique(numpy.array(dates).flatten())

There is surely a better way to gather and merge the matches, but you get the idea... You could also use numpy.argwhere((a < plus) * (a > minus)) to return the index instead of the date and use the index to grab the whole row and place it into your new array.

Collectives™ on Stack Overflow

Numpy array conditional matching

3 Answers 3

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related