How to find position of similar numbers in a list

python find position in list
python find index of all occurrences in list
python find index of value in numpy array
how to find the index of a character in a list in python
find index of element in array python
how to start list index from 1 in python
python select list elements by index
find index of element in 2d list python

Given a list of elements, with certain similar elements in it, how do I write a code in Python to find the position of all the similar elements in it? This must be done using a for loop and if conditions within the for loop.

list1 = [5, 90, 10, 5, 100, 5]

So in this case, since 5 is the element that has repeats or ties, the output will be 0,3,5.

You can try to use set to find the unique set of elements in the list and then check for indexes of the repeated elements.

Try the code below:

from collections import defaultdict

list1 = [5, 90, 10, 5, 100, 5]

set1 = set(list1)
res_dict = defaultdict(list)

for x in set1:
    for i, y in enumerate(list1):
        if x == y:
            res_dict[x].append(i)

print res_dict

Output:

{90: [1], 100: [4], 10: [2], 5: [0, 3, 5]}

Python List index(), If you provide a value for start , then end is optional. We shall look into examples, where we go through each of these scenarios in detail. Example 1: Find Index� Create a sorted list of your words and use the bisect module to identify the point in the sorted list where your word would fit according to the sorting order. Based on that position you can give the k nearest neighbours above and below to find the 2k closest words.

With enumerate inside list comprehension for 5,

>>> list1 = [5, 90, 10, 5, 100, 5]
>>> all_index = [i for i, j in enumerate(list1) if j == 5]
>>> all_index

Output:

[0, 3, 5]

With loop for all element,

list1 = [5, 90, 10, 5, 100, 5]
result = {}
for e in list1:
    result[e] = [i for i, j in enumerate(list1) if j == e]
print(result)

Output:

 {90: [1], 10: [2], 100: [4], 5: [0, 3, 5]}

Python – Get Index or Position of Element in List, We can achieve this task by iterating through the list and check for that value and just append the value index in new list and print that. This is� In this article, we've learned different ways of finding an element in a List, starting with quick existence checks and finishing with field-based searches. We also looked at the third-party libraries Google Guava and Apache Commons as alternatives to the Java 8 Streams API.

Using np.where will get you what you want:

import numpy as np

list1 = [5, 90, 10, 5, 100, 5]

data = np.array(list1)
unique = sorted(list(set(data)))

match = {}
for i in range(len(unique)):
    match[unique[i]] = np.where(data == unique[i])[0]

print(match)
>>> {5: array([0, 3, 5]), 10: array([2]), 90: array([1]), 100: array([4])}

Python, Delete elements from a Numpy Array by value or conditions in Python. No Comments Yet. Leave a Reply Cancel reply. Your� Consider a series of numbers composed of only digits 4 and 7. The first few numbers in the series are 4, 7, 44, 47, 74, 77, 444, .. etc. Given a number constructed by 4, 7 digits only, we need to find the position of this number in this series. Examples: Input : 7 Output : pos = 2 Input : 444 Output : pos = 7

Using only a for loop and if conditions:

input_list = [5, 90, 100, 5, 100, 5]

elements = {}
result = []

for i, e in enumerate(input_list):
    if e in elements:
        if not elements[e][1]:
            result.append(elements[e][0])
            elements[e][1] = True
        result.append(i)
    else:
        elements[e] = [i, False]

print(sorted(result))

Python: Find index of element in List (First, last or all occurrences , Filter key-value pairs in dictionary. Keep pairs whose value is greater than 1 i.e. only duplicate elements from list. dictOfElems = { key:value for� Example 2. Find Nth occurrence of a given character in a text string. Supposing you have some text strings in column A, say a list of SKUs, and you want to find the position of the second dash (-) in a string. The following formula works a treat: =FIND("-", A2, FIND("-",A2)+1) The first two arguments are easy to interpret: locate a dash

Python: Find duplicates in a list with frequency count & index positions, Originally Answered: How do I get the index value of a duplicate value in a list in Python? i tried the following code..it works fine.. original_values_list = [1,2,3,4,4� So, in our case it yields the indexes of ‘Ok’ in the list. We collected those indexes in a list. Find indexes of items in a list that satisfy certain conditions. Suppose we want to find indexes of items in a list that satisfy certain condition like indexes of items in a list of strings whose length is less than 3.

How to get the index of a duplicated string in a list (Python 2), A list is an ordered set of values, where each value is identified by an index. If you try to read or write an element that does not exist, you get a runtime error:. In general, you can find the K largest (or smallest) numbers in an array using a single pass for any K. The total time complexity will be O(NK), where N is the size of the array: Keep a sorted list of numbers that has at most K elements. Walk though the array and for each item: if there list is not yet full, insert the item

9. Lists — How to Think Like a Computer Scientist: Learning with , Use enumerate() to create a list of (index, value) tuples from the input list a_list . Use a list comprehension on this list of tuples to select every index that� If you want to use loops, you'll have to use a list or a set of numbers which you've already seen. Then while looping you'll check, with the in operator if the number is already seen. seen = [] for number in myList: if number in seen: print "Number repeated!"

Comments
  • Have you tried anything? Any possible approaches in mind?
  • Possible duplicate of Finding the index of an item given a list containing it in Python
  • "This must be done using a for loop and if conditions within the for loop." This is what we would like to tell you as well. Try and later come back with a specific question if you're stuck.