## Indices of equal elements

count equal element pairs in the given array

given an array we need to find that element whose value is equal to that of its index value

find all pairs in array

count number of pairs in array

find unique pair in an array with pairs of numbers

find all possible pairs in array

number of pairs in an array

**This question already has answers here**:

you can use list comprehension:

ones_indices = [i for i, e in enumerate(list_example) if e == 1]

**Index of the elements which are equal to the sum of all succeeding ,** The task is to find the index of the elements which are equal to the sum of all succeeding elements. If no such element exists then print -1. Examples: Input: arr [] = {� The idea is to count the frequency of each number and then find the number of pairs with equal elements. Suppose, a number x appears k times at index i 1, i 2,….,i k. Then pick any two indexes i x and i y which will be counted as 1 pair. Similarly, i y and i x can also be pair.

Without loops or imports you can do it this way:

list(zip(*list(filter(lambda x: x[1] == 1, enumerate(list_example)))))[0] # (0, 3, 7, 8)

or if you want lists:

list(map(list, zip(*list(filter(lambda x: x[1] == 1, enumerate(list_example))))))[0] # [0, 3, 7, 8]

**Indices of equal elements,** you can use list comprehension: ones_indices = [i for i, e in enumerate( list_example) if e == 1]. I am looking for a simple way to group the list-indices of equal elements of a list. Examples: {1,2,3,3} -> {{1}, {2}, {3,4}} (the elements at positions 1 and 2 are unique, and those at positio

my attempt so far:

g = [0,0,1,0,1,0,0,1,1,0,0,0,1,0] u = [] for i in g: if i == 1: u.append(g.index (i)) print (u) input ('continue')

**Output of indices of equal elements in an array,** Hello! I need to write a program that will display an element with max value and its index or indices in the array. Like this: : 2 : 3 : 7 : 9 : 2 : 9 Max� Usually, we require to find the index, in which the particular value is located. There are many method to achieve that, using index() etc. But sometimes require to find all the indices of a particular value in case it has multiple occurrences in list.

You could use `np.argwhere()`

to obtain incides in a list whose elements satisfy a certain condition.
Below's the code:

import numpy as np np.argwhere(np.array(list_example))[0]

Output:

[0, 3, 7, 8]

where we obtain indices of all non-zero elements in `list_example`

. For more specific condition checking, you could use,

np.argwhere(np.array(list_example==1))

**Finding the indices of the elements of one array in another ,** Given two vectors A and B, find the index, idx into A of the element of B so that. A( idx)=B. Now I know there must be many ways it can be done, but is there a� Find the nonzero elements in a 4-by-2-by-3 array. Specify two outputs, row and col, to return the row and column subscripts of the nonzero elements. When the input is a multidimensional array (N > 2), find returns col as a linear index over the N-1 trailing dimensions of X.

**Finding the indices of duplicate values in one array,** repeatedElements = values(counts >= 2). % Assume they're integers. % Print them out and collect indexes of repeated elements into an array. indexes = [];. If only condition argument is given then it returns the indices of the elements which are TRUE in bool numpy array returned by condition. For example following condition, boolArr = (arr == 15) returns a bool numpy array boolArr, containing TRUE of each element that is equal to 15, for other elements it contains False i.e.

**Simple way to group the list-indices of equal elements ,** Write a NumPy program to find indices of elements equal to zero in a NumPy array. Sample Solution: Python Code: import numpy as np nums =� size - 1, x, output); // If the element at index 0 is equal. // to x then add 1 to the array values. // and shift them right by 1 step. if (input [0] == x) {. for (int i = smallAns - 1; i >= 0; i--) {. output [i + 1] = output [i] + 1; } // Put the start index in front.

**NumPy: Find indices of elements equal to zero in a NumPy array ,** produce a vector whose four elements are all equal to 13. Similarly, by indexing a scalar with two vectors of ones it is possible to create a matrix. The following� The result is a list of all even elements in A that are less than 9. The use of the logical NOT operator, ~, converts the matrix mod(A,2) into a logical matrix, with a value of logical 1 (true) located where an element is evenly divisible by 2. Finally, find the elements in A that are less than 9 and even numbered and not equal to 2.

##### Comments

- What have you tried so far?
- Use a list comprehension and the
`enumerate()`

function. - Not that it necessarily can’t be done without loops but... why? Why is that the goal?
- If you consider a list comprehension to be a loop, my suggestion won't work. You can use
`enumerate`

,`filter`

, and`map`

to do it. - Four methods. The 4th uses filter so it does it using only built-in functions (i.e. without looping in user code).
- The condition is always 'Find the indices of array elements that are non-zero, grouped by element.' from docs.scipy.org/doc/numpy/reference/generated/… ;)