CODEWITHSUNDEEP
pythonnumpy
user2483176
user2483176

Reputation: 317

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 what I have tried, 

a = len(fs)
c = np.zeros(a)

for i in fs[0:a]:
    if i >= 0.75:
        print = 1

    elif i < 0.75:
        print = 0

fs is my column.

This code correctly prints the 0's and 1's I want, but I am unsure of how to count the number of 1's printed. I thought to first make an array of zeros, then somehow append the array in the loop to have an array of the correct 0's and 1's. Then I could just sum the array. I am not quite sure how to go about this and everything I try is not working (I am pretty inexperienced in programming). Does anyone have any advice about how to go about this? I know this is pretty simple...

Thank you!

Upvotes: 4

Views: 4123

Answers (3)

JoshAdel
JoshAdel

Reputation: 68682

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)

Upvotes: 3

icecrime
icecrime

Reputation: 76745

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

Upvotes: 1

vaggelas
vaggelas

Reputation: 68

If i understand correctly

you can use

countone = 0  #counter for the times you print one
countzero = 0 # counter fot the times you print 0

for 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?

Upvotes: -1

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