numpy: Counting in a 2D array where an element satisfies a condition

Question

I have a numpy array like below. I need a count of rows where the first element is 2. So in the array below, four rows start with 2 - the answer would be 4. How is this best accomplished in numpy? (I cannot use pandas, but can use scipy).

array([[1, 4, 5],
       [1, 4, 5],
       [2, 4, 5],
       [2, 4, 5],
       [2, 4, 5],
       [2, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5],
       [3, 4, 5]])

Answer 1

First, take the first column, all rows:

a[:,0]

Then, find the 2 s:

a[:,0] == 2

That gives you a boolean array. Which you can then sum:

(a[:,0] == 2).sum()

Answer 2

There is np.count_nonzero which in a common idiom is applied to logical arrays generated by evaluating a condition

np.count_nonzero(data[:, 0] == 2)

Btw. it's probably just for the sake of example, but if your array is sorted like yours you can also use np.searchsorted

np.diff(np.searchsorted(data[:, 0], (2, 3)))[0]

Answer 3

One more approach in addition to above approaches

>>> x[:,0]==2
array([False, False,  True,  True,  True,  True, False, False, False,
       False, False, False], dtype=bool)

will give you TRUE for the rows which have first column as 2.

>>> x[x[:,0]==2]
array([[2, 4, 5],
       [2, 4, 5],
       [2, 4, 5],
       [2, 4, 5]])

gives you corresponding rows and which satisfy the required condition. Now, you can use shape function to get length.

x[x[:,0]==2].shape[0]