Running maximum of numpy array values

Question

I need a fast way to keep a running maximum of a numpy array. For example, if my array was:

x = numpy.array([11,12,13,20,19,18,17,18,23,21])

I'd want:

numpy.array([11,12,13,20,20,20,20,20,23,23])

Obviously I could do this with a little loop:

def running_max(x):
    result = [x[0]]
    for val in x:
        if val > result[-1]:
            result.append(val)
        else:
            result.append(result[-1])
    return result

But my arrays have hundreds of thousands of entries and I need to call this many times. It seems like there's got to be a numpy trick to remove the loop, but I can't seem to find anything that will work. The alternative will be to write this as a C extension, but it seems like I'd be reinventing the wheel.

i would call that the cumulative max - running max suggests a window to me. unfortunately googling for that doesn't turn up anything useful. — andrew cooke
– andrew cooke, Commented Aug 31, 2011 at 0:54
i don't have numpy installed, but max.accumulate might work. check out "accumulate" in the docs. — andrew cooke
– andrew cooke, Commented Aug 31, 2011 at 0:56
@andrew max doesn't have an accumulate attribute in numpy. That would have been a good built-in solution though if it did. — JoshAdel
– JoshAdel, Commented Aug 31, 2011 at 1:01
tiny, tiny correction to the for loop, I think it should read for val in x[1:]: — Andras Vanyolos
– Andras Vanyolos, Commented May 13, 2024 at 13:01

Charles Beattie · Accepted Answer · 2011-08-31 01:16:33Z

83

numpy.maximum.accumulate works for me.

>>> import numpy
>>> numpy.maximum.accumulate(numpy.array([11,12,13,20,19,18,17,18,23,21]))
array([11, 12, 13, 20, 20, 20, 20, 20, 23, 23])

answered Aug 31, 2011 at 1:16

Charles Beattie

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1 Comment

Charles Beattie Over a year ago

wim got there just before I did.

Steve Tjoa · Accepted Answer · 2011-08-31 01:08:57Z

3

As suggested, there is scipy.maximum.accumulate:

In [9]: x
Out[9]: [1, 3, 2, 5, 4]

In [10]: scipy.maximum.accumulate(x)
Out[10]: array([1, 3, 3, 5, 5])

answered Aug 31, 2011 at 1:08

Steve Tjoa

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2 Comments

Robert Kern Over a year ago

There is no need to get it from the scipy namespace. It's a numpy ufunc. The duplication of the numpy symbols in scipy.* is a backwards-compatibility leftover from the days of Numeric.

Steve Tjoa Over a year ago

Sorry about that. Personal bias, I guess.

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