How to count values in a certain range in a Numpy array?

问题:

I have a NumPy array of values. I want to count how many of these values are in a specific range say x25. I have read about the counter, but it seems to only be valid for specif values not ranges of values. I have searched, but have not found anything regarding my specific problem. If someone could point me towards the proper documentation I would appreciate it. Thank you

I have tried this

   X = array(X)    for X in range(25, 100):        print(X) 

But it just gives me the numbers in between 25 and 99.

EDIT The data I am using was created by another program. I then used a script to read the data and store it as a list. I then took the list and turned it in to an array using array(r).

Edit

The result of running

 >>> a[0:10]  array(['29.63827346', '40.61488812', '25.48300065', '26.22910525',    '42.41172923', '20.15013315', '34.95323355', '13.03604098',    '29.71097606', '9.53222141'],    dtype='

回答1:

If your array is called a, the number of elements fulfilling 25 is

((25 

The expression (25 results in a Boolean array with the same shape as a with the value True for all elements that satisfy the condition. Summing over this Boolean array treats True values as 1 and False values as 0.

回答2:

You could use histogram. Here's a basic usage example:

>>> import numpy >>> a = numpy.random.random(size=100) * 100  >>> numpy.histogram(a, bins=(0.0, 7.3, 22.4, 55.5, 77, 79, 98, 100)) (array([ 8, 14, 34, 31,  0, 12,  1]),   array([   0. ,    7.3,   22.4,   55.5,   77. ,   79. ,   98. ,  100. ])) 

In your particular case, it would look something like this:

>>> numpy.histogram(a, bins=(25, 100)) (array([73]), array([ 25, 100])) 

Additionally, when you have a list of strings, you have to explicitly specify the type, so that numpy knows to produce an array of floats instead of a list of strings.

>>> strings = [str(i) for i in range(10)] >>> numpy.array(strings) array(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'],        dtype='|S1') >>> numpy.array(strings, dtype=float) array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.]) 

回答3:

Building on Sven's good approach, you can also do the more direct:

numpy.count_nonzero((25 

This first creates an array of booleans with one boolean for each input number in array a, and then count the number of non-False (i.e. True) values (which gives the number of matching numbers).

Note, however, that this approach is twice as slow as Sven's .sum()

回答4:

Sven's answer is the way to do it if you don't wish to further process matching values.
The following two examples return copies with only the matching values:

np.compress((25 

Or:

a[(25 

Example interpreter session:

>>> import numpy as np >>> a = np.random.randint(200,size=100) >>> a array([194, 131,  10, 100, 199, 123,  36,  14,  52, 195, 114, 181, 138,        144,  70, 185, 127,  52,  41, 126, 159,  39,  68, 118, 124, 119,         45, 161,  66,  29, 179, 194, 145, 163, 190, 150, 186,  25,  61,        187,   0,  69,  87,  20, 192,  18, 147,  53,  40, 113, 193, 178,        104, 170, 133,  69,  61,  48,  84, 121,  13,  49,  11,  29, 136,        141,  64,  22, 111, 162, 107,  33, 130,  11,  22, 167, 157,  99,         59,  12,  70, 154,  44,  45, 110, 180, 116,  56, 136,  54, 139,         26,  77, 128,  55, 143, 133, 137,   3,  83]) >>> np.compress((25 >> a[(25 

The above examples use a "bit-wise and" (&) to do an element-wise computation along the two boolean arrays which you create for comparison purposes.
Another way to write Sven's excellent answer, for example, is:

np.bitwise_and(25 

The boolean arrays contain True values when the condition matches, and False when it doesn't.
A bonus aspect of boolean values is that True is equivalent to 1 and False to 0.

回答5:

I think @Sven Marnach answer is quite nice, because it operates in on the numpy array itself which will be fast and efficient (C implementation).

I like to put the test into one condition like 25 , so I would probably do it something like this:

len([x for x in a.ravel() if 25

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