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Home/ Questions/Q 8856103
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T14:17:04+00:00 2026-06-14T14:17:04+00:00

first question here, so i will get right to it: using python 2.7 I

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first question here, so i will get right to it:

using python 2.7

I have a dictionary of items, the keys are an x,y coordinate represented as a tuple: (x,y) and all the values are Boolean values.

I am trying to figure out a quick and clean method of getting a count of how many items have a given value. I do NOT need to know which keys have the given value, just how many.

there is a similar post here:
How many items in a dictionary share the same value in Python, however I do not need a dictionary returned, just an integer.

My first thought is to iterate over the items and test each one while keeping a count of each True value or something. I am just wondering, since I am still new to python and don’t know all the libraries, if there is a better/faster/simpler way to do this.

thanks in advance.

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  1. Editorial Team
    Editorial Team
    2026-06-14T14:17:05+00:00Added an answer on June 14, 2026 at 2:17 pm

    This first part is mostly for fun — I probably wouldn’t use it in my code.

    sum(d.values())
    

    will get the number of True values. (Of course, you can get the number of False values by len(d) - sum(d.values())).


    Slightly more generally, you can do something like:

    sum(1 for x in d.values() if some_condition(x))
    

    In this case, if x works just fine in place of if some_condition(x) and is what most people would use in real-world code)

    OF THE THREE SOLUTIONS I HAVE POSTED HERE, THE ABOVE IS THE MOST IDIOMATIC AND IS THE ONE I WOULD RECOMMEND


    Finally, I suppose this could be written a little more cleverly:

    sum( x == chosen_value for x in d.values() )
    

    This is in the same vein as my first (fun) solution as it relies on the fact that True + True == 2. Clever isn’t always better. I think most people would consider this version to be a little more obscure than the one above (and therefore worse).

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