# How to find a median of odd and even numbers from a list of numbers

For a given list, say {94,35,47,29,86,40,46,72,87}, how can I find a list of even and odd numbers.

Here is the output after trying with the suggested solution in the answer.

After putting the number in list as told by @Nasser, I got the intended output.

I am seeking to solve this problem:

Create an Association that lists the Median of the even and odd numbers respectively under the keys True and False. (Hint: Use GroupBy with EvenQ.)

By applying GroupBy, I could now list the even and odd numbers:

Input:

GroupBy[{94,35,47,29,86,40,46,72,87,84,49,82,100,99,7,70,67,63,13,
91,18,94,94,15,72,39,4,7,31,67,65,37,93,9,4,84,97,60,48,61,91,24,10,
46,42,28,41,36,62,70,78,38,80,70,63,20,21,7,54,26,17,73,100,56,36,
54,100,46,0,59,77,80,57,82,63,74,8,47,68,74,54,57,4,90,14,31,88,23,
30,37,95,44,100,12,23,48,4,14,100,57},EvenQ]


Output:

True->  {94,86,40,46,72,84,82,100,70,18,94,94,72,4,4,84,60,48,24,10,46,
42,28,36,62,70,78,38,80,70,20,54,26,100,56,36,54,100,46,0,80,
82,74,8,68,74,54,4,90,14,88,30,44,100,12,48,4,14,100},
False->  {35,47,29,87,49,99,7,67,63,13,91,15,39,7,31,67,65,37,93,9,97,
61,91,41,63,21,7,17,73,59,77,57,63,47,57,31,23,37,95,23,57}


What is needed is the count. True->54, False->49.

With CountsBy as suggested by @sjoerd-smit, I get the count of even and odd numbers in one command.

CountsBy[{94,35,47,29,86,40,46,72,87,84,49,82,100,99,7,70,67,63,13,
91,18,94,94,15,72,39,4,7,31,67,65,37,93,9,4,84,97,60,48,61,91,24,10,
46,42,28,41,36,62,70,78,38,80,70,63,20,21,7,54,26,17,73,100,56,36,
54,100,46,0,59,77,80,57,82,63,74,8,47,68,74,54,57,4,90,14,31,88,23,
30,37,95,44,100,12,23,48,4,14,100,57},EvenQ]


By Median applied individually, I could get the medians of even and odd numbers in the list.

Input:

Median[{35,47,29,87,49,99,7,67,63,13,91,15,39,7,31,67,65,37,93,
9,97,61,91,41,63,21,7,17,73,59,77,57,63,47,57,31,23,37,95,23,57}]


Output: 49

Similarly, for the list of even numbers.

Might be a way to get the same with one command.

• Please take a look at GatherBy
– Syed
Sep 3 at 6:36
• @Syed I recommend always using GroupBy over GatherBy, to be honest. It's just a more powerful function and GatherBy can be unpredictable in the ordering of the sublists. Sep 3 at 8:05
• Dear OP, please start a fresh question for new queries, instead of morphing the question.
– Syed
Sep 3 at 8:40
• @Syed Did not start a new question as from my previous experience, I was told to remove the first question as I formulated the second question based on input from the first one. Sep 4 at 5:43
• The problem I have is that I am no longer sure what you are asking. But maybe you want GroupBy[{94,35,...},EvenQ,Median] ( <|True -> 54, False -> 49|> ) or if you want both Counts and Median, maybe GroupBy[{94,35,...},EvenQ,#]&/@{Length,Median} ( {<|True -> 59, False -> 41|>, <|True -> 54, False -> 49|>} ). But perhaps you require something else? Sep 4 at 8:21

If you only need the counts, then CountsBy is the way to go:

CountsBy[Range[100], EvenQ]


You can also do it with the 3rd argument of GroupBy:

GroupBy[Range[100], EvenQ, Length]


If you want the results in a list in a particular order (say, even then odd), you can use Lookup:

Lookup[CountsBy[Range[100], EvenQ], {True, False}, 0]


I think this was asked before, but I can't find it now searching.

lis = {94, 35, 47, 29, 86, 40, 46, 72, 87}
{Select[lis, EvenQ[#] &], Select[lis, OddQ[#] &]}


I am sure there are many other ways to do this.

edit:

A little simpler version as suggested in comment thanks to kglr is

{Select[lis, EvenQ], Select[lis, OddQ]}

• you can use simpler Select[lis, EvenQ]
– kglr
Sep 3 at 6:54
• Getting the first few elements in black (attached image with the question). Is it due to some mismatch with global variables? Sep 3 at 7:21
• @SplendidDigitalSolutions you need to put the numbers in a list first. a list is { ....} then it will work. You can't just list the number as you did. Sep 3 at 7:28

Try Cases:

list={94, 35, 47, 29, 86, 40, 46, 72, 87};
twolists=List[Cases[list, Except[_?EvenQ]], Cases[list, Except[_?OddQ]]]
(*{{35, 47, 29, 87}, {94, 86, 40, 46, 72}}*)
Map[Median, twolists]
(*{41,72}*)


With order:

List[Sort[Cases[list, Except[_?EvenQ]], Less], Sort[Cases[list, Except[_?OddQ]], Less]]
(*{{29, 35, 47, 87}, {40, 46, 72, 86, 94}}*)
List[Sort[Cases[list, Except[_?EvenQ]], Greater], Sort[Cases[list, Except[_?OddQ]], Greater]]
(*{{87, 47, 35, 29}, {94, 86, 72, 46, 40}}*)