## IF statement for repeating elements

I have interesting question:
Let's assume that we have `lst = [1,2,3,4,5,5,5,5,5,6,7,8]`

I am interested on saying to if statement only return
result of 3rd 5.
for example:

for i in range(0,lent(lst)): if lst[i]==5: print(i,"is index of 1st 5") break

but how to say ask if to show 3rd 5's index without additional parameter or listing?

This should fit your needs

counter = 0 for item in lst: if item == 5: counter += 1 if counter >= 3: break

**Repetition Statements,** A repetition statement is used to repeat a group (block) of programming instructions. If the for loop will process all of the elements in the list the same, use the� IF statement for repeating elements. Ask Question Asked 8 months ago. Active 8 months ago. Viewed 49 times 2. I have

You can use enumerate on the list and get all the indices of the required element then you get whatever index you want

lst = [1,2,3,4,5,5,5,5,5,6,7,8] occurances = [idx for idx, val in lst if val == 5] occurances[0] # first occurance occurances[1] # second occurance occurances[2] # third occurance

**Repeat/Until Block Loops,** Since the condition is evaluated at the end of each iteration, a repeat / until loop will always be executed at least once, even if the condition is already true when� The for-each statement allows you to loop through repeating elements. To repeat elements with the for-each statement: Drag a source to a target to create a simple mapping.

juanpa.arrivillaga already pointed the usual solution out. However, with e.g. numpy you could do also another way:

import numpy as np lst = np.array([1,2,3,4,5,5,5,5,5,6,7,8]) cnt5 = (lst==5).cumsum() np.arange(len(lst))[cnt5==3][0] # 6

cnt5 is the cumulative sum of the boolean array, which is True only at indices where lst has a `5`

, i.e.: it's an array of the counters of lst's 5's.
Then you only need the index where this array is `3`

.

**Find the two repeating elements in a given array,** cout << " Repeating elements are " ;. for (i = 0; i < size; i++). {. if (count[arr[i]] == 1). cout << arr[i] << " " ;. else. count[arr[i]]++;. } } // Driver code. You are given an array of n+2 elements.All elements of the array are in range 1 to n.All elements occur once except two numbers, which occur twice. Your task is to find the two repeating numbers.

**Chapter 3: Logicals and Loops,** Tests if all elements in vector satisfy both conditions and returns a single There are three loop functions: * Repeat * While * For Tools help shape loops to� Let the repeating numbers be X and Y, if we xor all the elements in the array and all integers from 1 to n, then the result is X xor Y.

**Repetition Statements, Arrays and Structured Programming ,** More commonly, a section of code that is repeated is referred to as a loop, because The first necessary element is a repetition statement. If the test occurs at the beginning of the loop, the type of loop is called a pre-test� Loop through the matrix and assign each element a new value. Assign 2 on the main diagonal, -1 on the adjacent diagonals, and 0 everywhere else. for c = 1:ncols for r = 1:nrows if r == c A(r,c) = 2; elseif abs(r-c) == 1 A(r,c) = -1; else A(r,c) = 0; end end end A

Conditional rendering in React works the same way conditions work in JavaScript. Use JavaScript operators like if or the conditional operator to create elements representing the current state, and let React update the UI to match them. Consider these two components:

##### Comments

- Also, take a look at meta.stackexchange.com/questions/22186/…
- Anyway, here you could just keep a
*count*fo the number of five's you've seen, and then break on the condition`list[i] == 5 and count == 2`

. Note, that means you are seeing the 3rd 5, since you've already counted 2 - Duplicate of stackoverflow.com/questions/22267241/…
- If I told you the list of numbers one at a time, how would you solve this problem by hand? You should describe these steps
**in words**using whatever written language you know best. Once you have a good understanding of how to solve the problem yourself, it will be easier to tell the computer how to do it.