In 2.7 and 3.1, there is the special Counter (dict subclass) for this purpose.
>>> from collections import Counter
>>> Counter(['apple','red','apple','red','red','pear'])
Counter({'red': 3, 'apple': 2, 'pear': 1})
Top answer 1 of 10
420
In 2.7 and 3.1, there is the special Counter (dict subclass) for this purpose.
>>> from collections import Counter
>>> Counter(['apple','red','apple','red','red','pear'])
Counter({'red': 3, 'apple': 2, 'pear': 1})
2 of 10
363
I like:
counts = dict()
for i in items:
counts[i] = counts.get(i, 0) + 1
.get allows you to specify a default value if the key does not exist.
GeeksforGeeks
geeksforgeeks.org › python › count-dictionaries-in-a-list-in-python
Count dictionaries in a list in Python - GeeksforGeeks
July 12, 2025 - The program only creates a single list comprehension and a few variables to store the input list and the result. Method #2: Using recursion + isinstance() ( for nested dictionaries) The combination of above functionalities can be used to solve this problem. In this, we also solve the problem of inner nesting using recursion. ... # Python3 code to demonstrate working of # Dictionary Count in List # Using recursion + isinstance() # helper_func def hlper_fnc(test_list): count = 0 if isinstance(test_list, str): return 0 if isinstance(test_list, dict): return hlper_fnc(test_list.values()) + hlper_f
Reddit
reddit.com › r/learnpython › conversion between list and dictionary
r/learnpython on Reddit: conversion between list and dictionary
November 1, 2021 - It's O(n2) in the worst case (~N unique keys => loop through entire list N times to count occurrences). The set conversion in that code saves a lot of time if you only have a couple unique keys. ... If you want to use comprehension you can use set to create a unique list. dict = {element:lst.count(element) for element in list(set(lst))}
Built In
builtin.com › software-engineering-perspectives › convert-list-to-dictionary-python
10 Ways to Convert Lists to Dictionaries in Python | Built In
Converting a list to a dictionary with the same value for all keys. Converting a list to a dictionary using dict.fromkeys(). Converting a nested list to a dictionary using dictionary comprehension. Converting a list to a dictionary using Counter(). An introduction on how to convert a list to dictionary in Python...
Reddit
reddit.com › r/learnpython › how do i count a certain key value in a list python
r/learnpython on Reddit: how do I count a certain key value in a list python
January 24, 2022 -
I am trying to count how many times "yes" value appears in the done key but every time i rune the script i end up with a 0 can someone help please ?
list=[{"task":"walk dog", "done":"yes"},
{"task":"talk", "done":"yes"},
{"task":"sleep", "done":"yes"},
{"task":"eat", "done":"no"},
{"task":"chicken", "done":"yes"},
{"task":"smoke", "done":"yes"}
]
c = 0
for item in list:
if list[1] == "yes":
c= c+1
print(c) Top answer 1 of 6
4
done worry i fixed my own code. in case anyone was interested or for the future list=[{"task":"walk dog", "done":"yes"}, {"task":"talk", "done":"yes"}, {"task":"sleep", "done":"yes"}, {"task":"eat", "done":"no"}, {"task":"chicken", "done":"yes"}, {"task":"smoke", "done":"yes"} ] c = 0 for item in list: if item["done"] == "yes": c= c+1 print(c)
2 of 6
3
You could use a shortcut for this as well, c = sum(d['done'] == "yes" for d in data) # don't use list as variable name as the boolean values True and False are a subclass of int and equal to 1 and 0 respectively. This uses a generator expression (similar to list comprehension, well documented) that loops through the list and checks for the condition you want to count. That generator expression is passed to sum to add up the numbers. Note that if you had a dictionary missing the 'done' key, you would get an error. You could use the dict.get method to avoid this.
DevQA
devqa.io › count-occurance-elements-list-python
How to Count the Occurrences of Each Element in a List using Python
July 8, 2023 - In this approach, we iterate through each element of the list and check if it already exists as a key in the count_dict. If it does, we increment its value by 1; otherwise, we add the element as a key with an initial value of 1. Finally, we ...
Stack Overflow
stackoverflow.com › questions › 58350482 › how-do-i-convert-list-with-word-count-into-a-dictionary
python - How do i convert list with word count into a dictionary - Stack Overflow
s = 'I want to make a dictionary ... 1 · Again we can use Counter which is faster. from collections import Counter s = list(s) print(Counter(s))...
Reddit
reddit.com › r/learnpython › how to get count of unique dictionaries in list?
r/learnpython on Reddit: How to get count of unique dictionaries in list?
December 23, 2020 -
Hi,
let's assume I have the following list:
a = [{'key': 'value1', 'key2': 'value2'},
{'key': 'value1', 'key2': 'value2'},
{'key': 'value3', 'key4': 'value5'}]
I would like to now get the count of each dictionary in this list, as a key: value in the dict.
Example output:
[{'key': 'value3', 'key4': 'value5', 'count': 1},
{'key': 'value1', 'key2': 'value2', 'count': 2}]Is there are a smarter way than some sort of loop? It should work but I would like to keep my O() down as much as possible.
I know that there are some examples for lists of strings or integers but they are all hashable types which dictionaries are not.
Top answer 1 of 4
3
I'm sorry but I can't test out code right now or build one really, but right off the top of my head I would suggest creating a condensed the given list using set() so you can grab the unique entries. From there, I would normally just use a list comprehension that returns an element from the set but updated with .update({'count':a.count(elem)}). The main reason why I would rather have to test this rather than build a code as an answer is that dictionaries are quite finnicky when it comes to copies, so there might be some deepcopying that needs to be done along the way. But normally, yes, the way i do it is make a copy of the list using set(a) and use a list comprehension that involves using a.count() for each element in the set.
2 of 4
2
Not saying it doesn't exist, but I don't know of an efficient way to do this outside of a loop. Some sort of iteration needs to be done to check these values and compare them with the others. I'm curious what solutions others may have. RemindMe! 5 hours
GeeksforGeeks
geeksforgeeks.org › python › counting-the-frequencies-in-a-list-using-dictionary-in-python
Counting the Frequencies in a List Using Dictionary in Python - GeeksforGeeks
October 25, 2025 - The defaultdict from the collections module automatically initializes new keys with a default value (in this case, 0), so there's no need to explicitly check if the key exists before incrementing the count. ... from collections import defaultdict a = ['apple', 'banana', 'apple', 'orange', 'banana', 'banana'] freq = defaultdict(int) for item in a: freq[item] += 1 print(dict(freq)) ... For each item in a, freq[item] += 1 increments its count. This will count the frequency of each item in the list using get() method.
w3resource
w3resource.com › python-exercises › dictionary › python-data-type-dictionary-exercise-34.php
Python: Count number of items in a dictionary value that is a list - w3resource
Write a Python program to count the number of items in a dictionary value that is a list. ... # Create a dictionary 'dict' with names as keys and lists of subjects as values. dict = {'Alex': ['subj1', 'subj2', 'subj3'], 'David': ['subj1', 'subj2']} # Calculate the total number of subjects by summing the lengths of the lists (values) in the dictionary.
Data Science Dojo
discuss.datasciencedojo.com › python
How to use a Python dictionary for counting? - Python - Data Science Dojo Discussions
May 11, 2023 - Hello everyone! I’m working on text analysis in Python and have stored my data from txt files in Python lists. Now, I need to count the frequency of words in the data. For instance, given this sample list: data_list = […
Python Guides
pythonguides.com › python-dictionary-count
Count Occurrences in Python Dictionary
January 12, 2026 - It is very helpful when you are building a complex Python dictionary count system. Unlike a regular Python dictionary, defaultdict automatically assigns a default value to a new key. from collections import defaultdict # List of professional US sports leagues leagues = ["NFL", "NBA", "MLB", "NFL", "NHL", "NBA", "NFL"] # Initialize defaultdict with int (which defaults to 0) league_counts = defaultdict(int) for league in leagues: league_counts[league] += 1 print(dict(league_counts))
Medium
medium.com › @dxsmith12 › efficiently-generating-a-python-dictionary-to-store-item-counts-a00e365767ef
Efficiently Generating a Python Dictionary to Store Item Counts | by Darren Smith | Medium
January 24, 2022 - import collections d = collections.defaultdict(int) sentence = 'the quick brown fox jumps over the lazy dog' for word in sentence.split(' '): d[word] += 1 print(d.items()) ... dict_items([('the', 2), ('quick', 1), ('brown', 1), ('fox', 1), ('jumps', 1), ('over', 1), ('lazy', 1), ('dog', 1)]) Lastly and most concisely is the use of the collections module’s Counter class. In the example below a Counter object may be created with letters (keys) and number of occurrences (values) contained within the list. Like the defaultdict, the Counter object is a subclass of the standard python dictionary and may be interacted with as if it was a standard dictionary.
Team Treehouse
teamtreehouse.com › community › counting-the-number-of-list-items-in-a-dictionarys-values
Counting the number of list items in a dictionary's values (Example) | Treehouse Community
December 2, 2017 - count_dict = {k: len(v) for k,v in d.items() if len(v) > 3 } The above would filter the return dict to keys whose list length is greater than 3, etc... Posting to the forum is only allowed for members with active accounts.
Educative
educative.io › answers › how-to-count-the-number-of-occurrences-of-a-list-item-in-python
How to count the number of occurrences of a list item in Python
The counter method is a collection where elements are stored as a dictionary with keys and counts as values. It takes one argument. ... Let’s look at an example of this. l = ['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', ...
GeeksforGeeks
geeksforgeeks.org › python › python-convert-a-list-to-dictionary
Convert a List to Dictionary Python - GeeksforGeeks
For example, we are given a list a=[10,20,30] we need to convert the list in dictionary so that the output should be a dictionary like {0: 10, 1: 20, 2: 30}. We can use methods like enumerate, zip to convert a list to dictionary in python.
Published July 12, 2025
Top answer 1 of 4
3
Problem with your code is that you seem to iterate on all letters of the word. letter[0] is a substring of the letter (which is a string).
You'd have to do it more simply, no need for a double loop, take each first letter of your words:
for word in text:
if word: # to filter out empty strings
first_letter = word[0]
But once again collections.Counter taking a generator comprehension to extract first letter is the best choice and one-liner (with an added condition to filter out empty strings):
import collections
c = collections.Counter(x[0] for x in ["banana","ball", "cat", "", "hat"] if x)
c is now a dict: Counter({'b': 2, 'h': 1, 'c': 1})
one variant to insert None instead of filtering out empty values would be:
c = collections.Counter(x[0] if x else None for x in ["banana","ball", "cat", "", "hat"])
2 of 4
1
my_list=["banana","ball", "cat", "hat"]
my_dict = dict()
for word in my_list:
try:
my_dict[word[0]] += 1
except KeyError:
my_dict[word[0]] = 1
This increases the value of the key by 1 for the already existing key, and if they key has not been found before it creates it with the value 1
Alternative:
my_list=["banana","ball", "bubbles", "cat", "hat"]
my_dict = dict()
for word in my_list:
if word[0] in my_dict.keys():
my_dict[word[0]] += 1
else:
my_dict[word[0]] = 1
Stanford
web.stanford.edu › class › archive › cs › cs106a › cs106a.1204 › handouts › lecture-18.html
Python "dict" type
The possible keys are 'breakfast', 'lunch', 'dinner', although a key may or not be present in the meals dict. bad_start() - check for bad breakfast - return True if no breakfast or if it is 'candy' ... def bad_start(meals): if 'breakfast' not in meals: return True if meals['breakfast'] == 'candy': return True return False # Can be written with "or" / short-circuiting avoids key-error # if 'breakfast' not in meals or meals['breakfast'] == 'candy': ... Counts dict: key for each distinct value value for each key is count how many times that key appears
Top answer 1 of 3
2
if you have in your list only dictionaries than you can use the buid-in function len:
len(my_list)
if you have also other objects than dictionaries in your list than you can filter your list to keep only the dictionaries in a list comprehension than use the same len method:
len([e for e in my_list if isinstance(e, dict)])
or you can use the buit-in funcion sum:
sum(1 for e in my_list if isinstance(e, dict))
2 of 3
2
Try this:
dataList = [[0, {"1":1}, {"5":5}], {'a': 1,'b':2,'c':3}, {'a':3,'b': 3,'c':5}, {'a': 5,'b': 4,'c': 5}, [2, [2, {"100":2, "120":{100:1}}]]]
def numDict(li):
count = 0
if isinstance(li, str):
return 0
if isinstance(li, dict):
return return numDict(li.values()) + numDict(li.keys()) + 1
try:
for i in li:
count = count + numDict(i)
except TypeError:
return 0
return count
print(numDict(dataList))
It handles nested dictionaries too. I am using recursive and it has a limit on the number of recursive call it can make. If you exceed a certain depth of nested dictionaries, then you will get following error:
RecursionError: maximum recursion depth exceeded while calling a Python object
Output:
7
Studytonight
studytonight.com › python-programs › counting-the-frequencies-in-a-list-using-dictionary-in-python
Counting the frequencies in a list using dictionary in Python - Studytonight
We will define a function that accepts the list as a parameter. Then we will create a dictionary where the list element is the key and its frequency will be the value. To get frequency we will traverse the list and check if the element is already present in the dictionary or not and keep a count accordingly...