I am trying to code a numpy function that shows the number of values that are less than 0 within an array. How would I be able to do that?
import numpy as np
array = np.array([10,4,3,5,6,67,3,-12,5,-6,-7])
Expected Output:
3
2 Answers
Just compare the array against a value (i.e. less than zero for negative values), then call sum since, you only need the count.
>>> array = np.array([10,4,3,5,6,67,3,-12,5,-6,-7])
>>> (array<0).sum()
3
And if you want those values instead of just count, then use the masking to get those values
>>> array[array<0]
array([-12, -6, -7])
2 Comments
Robert Cotterman
i never realized lists did this, is this just a numpy array thing?
ThePyGuy
Yeah, it's numpy
you can print it with:
print(sum([i for i in array if i < 0]))
you can set it to a variable with:
lessThanZero = sum([i for i in array if i < 0])
did a benchmark (of sorts)
import datetime
import numpy
numpySum=0
forLoop=0
test = numpy.array([1,2,3,4,5,6,7,8,9,0])
for k in range(1000):
a = datetime.datetime.now()
for j in range(100):
z = (test<5).sum()
b = datetime.datetime.now() - a
a = datetime.datetime.now()
for j in range(100):
z = [i for i in test if i < 5]
a = datetime.datetime.now() - a
if b<a:numpySum+=1
else: forLoop+=1
print(numpySum, forLoop)
output over 3 runs:
633 367
752 248
714 286
appears the numpy sum is faster :D
This was for an array of size 10. I tried again with size 1000, and the numpy sum answer won every time!
3 Comments
hilberts_drinking_problem
This will give the sum of negative values instead of the count.
roganjosh
This is not the way to work with numpy arrays. This will run in python time, but a native numpy approach will run in C time (which could be orders of magnitude faster)
Robert Cotterman
I don't use numpy a lot, so I wanted to test it. Thank you @roganjosh that makes total sense. numpy clearly won.
(array < 0).sum().