# Manipulating List?

I have a huge list with 1000 elements. Manually entering and calculating SUM of probabilities it would take a few hours.

Example:

 list = {1, 2, 3, 3, 4, 3}


Calculating probabilities:

Counts[list]/Length[list]

<|1 -> 1/6, 2 -> 1/6, 3 -> 1/2, 4 -> 1/6|>


Calculating the sum of the list elements * probabilities

1*(1/6) + 2*(1/6) + 3*(1/2) + 3*(1/2) + 4*(1/6) + 3*(1/2)= 17/3


Could you give me some tips please?

Updated as I miss read the question

list = {1, 2, 3, 3, 4, 3};
counts= Counts[list]
(* <|1 -> 1, 2 -> 1, 3 -> 3, 4 -> 1|> *)


Keys[counts]*vals gives you number of occurrences times the value of the occurrence. Then calculate the product with Times on making a pair with this value and its probability. Finally, Total the list of results from the multiplication.

With[{vals = Values[counts]},
Total[Times[Sequence @@ #] & /@
Transpose@{Keys[counts]*vals, vals/Length[list]}]
]


Hope this helps.

• Yours answer is Wrong! Sum is 17/3 not 8/3...... Apr 25, 2015 at 15:13
• @Mariusz Updated as I misread the question. Apr 25, 2015 at 17:21
• Yours answer is correct and very speedly.:-) Apr 25, 2015 at 18:28

You can use Count and Dot to get 17/3 from your list:

list = {1, 2, 3, 3, 4, 3};
Dot[list, Count[list,#]/Length@list&/@list]
(* 17/3 *)

• Yours answer is correct :-) Apr 25, 2015 at 18:28

It is easier to see what is happening if you use symbols rather than numbers.

list = {a, b, c, c, d, c};

prob = Counts[list]/Length[list];


The sum of the product of the elements and their corresponding probabilities is given by either the Dot product

elem = Union[list];

elem.prob /@ elem // Simplify


1/6 (a + b + 3 c + d)

or more simply the Mean

Mean[list]


1/6 (a + b + 3 c + d)

If you are going to include each individual element in the sum then the probability of each individual element is just 1/Length[list]

Total[list/Length[list]] // Simplify


1/6 (a + b + 3 c + d)

It is unclear what statistic your calculation represents or to what purpose it would be used.

• .Yours answer is Wrong! Sum is 17/3 not 8/3. Apr 25, 2015 at 15:12

As @Bob said "it is unclear what statistic your calculation represents" ... but maybe you'll be interested in more explicit syntax (see also these docs):

list = {1, 2, 3, 3, 4, 3};


then for example, the probability to get 3 is :

Probability[x == 3, Distributed[x, list]]


1/2

Then what you ask for is :

list.(Probability[x == #, Distributed[x, list]] & /@ list)


17/3

but it looked like you were asking for :

Expectation[x, Distributed[x, list]]


8/3

• I wanted to sum ​​it got it 17/3.English is not my native language. While translating I had something to mess with the question.Google Translator ;-) May 1, 2015 at 9:17