Want To Count The Number Of Values In A Column That Meet A Condition

I am trying to count the number of values in a column that meet a certain condition (for example, are greater than 0.75). The column I have is made up of 2000+ decimals. This is w

Solution 1:

In numpy you could do something like:

np.where(fs >= 0.75)[0].size

or

np.count_nonzero(fs >= 0.75)

Both are equivalent, but I probably prefer the second. See the docs for an explanation:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.count_nonzero.html

but basically fs >= 0.75 creates a boolean array of the same length of fs where its elements are True or False based on the conditional. Since this is equivalent to 1 and 0 respectively, np.count_nonzero then returns a count of the non zero elements.

You can, of course, slice fs as well:

np.count_nonzero(fs[0:a] >= 0.75)

Solution 2:

It is not clear if you want to compute the number of 1 in the same loop, in which case vaggelas answer is correct.

If you want a separate loop to count the number of values >= 0.75, you can use:

>>>sum(1for i in fs[0:a] if i >= 0.75)

Solution 3:

If i understand correctly

you can use

countone = 0  #counter for the times you print one
countzero = 0 # counter fot the times you print 0for i in fs[0:a]:
    if i >= 0.75:
        print = 1
        countone+=1

    elif i < 0.75:
        print = 0 
        countzero +=1

that's what you meant?