### Thread: Calculating the statistical mode

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#### Calculating the statistical mode

how do i find the mode of a list?

by this i mean the value that appears the most in a particular list.

for example:
List (array): 1, 2, 3, 3, 3, 5, 5, 10
Mode : 3

or

List (array): 1, 2, 3, 3, 3, 5, 5, 5
Mode: 3 and 5

thank you.
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:rolleyes:
Perhaps you should create a list (i mean the data structure where every element has a pointer to the next element) which will contain two numbers - first is the value of an item in your original list, and second is the number of times you have encountered that value in your original list.

I hope my bad English and too complex phrase construction will not be a barrier between out developing minds %-[]

:rolleyes:
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as for your example - you'll have 4 nodes in your generated list, with values
1 1
2 1
3 3
5 3
respectively (pointers included).
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!

I wrote a program that found the mean, median, mode and range =) here is the part where it finds the mode:

Code:
```void mode(double* nums, unsigned int choice)
{
unsigned int index;
int count=0;
int max =0;
for(index=0; index<choice; index++)
{
if(nums[index]==nums[index+1])
count++;
else
{
if(count>max)
max=count;
count=0;
}
}
count=0;
for(index=0; index<choice; index++)
{
if(nums[index]==nums[index+1])
count++;
else
{
if(count==max)
cout<<"Mode(s):\t" <<nums[index]<<"\n";
count=0;
}
}
}```
:] hope this helps
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#### nonono

It will return only the count of the longest sequence of identical numbers.
But as I understand, you need to return the VALUE which is the most frequently appears in the whole sequence.
e.g. for a sequence

1, 2, 3, 4, 2, 4, 4, 2

your function will return THE COUNT 2 (of a value 4)
but the modes are THE VALUES 2 and 4 (appear 3 times both)
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i am not really sure what u are tryin to say :(