I have a two dimensional list.
[[3, 3, 5], [3, 2, 8], [2, 1, 3]] I want to count how many times 3 appears as the first value of each list within this list, preferably without iterating through.
- 4What do you mean "without iterating the list"? Do you want to avoid for loops and list comprehensions for some reason?Bill Lynch– Bill Lynch2021-01-04 04:34:21 +00:00Commented Jan 4, 2021 at 4:34
- Does this answer your question? How do I flatten a list of lists/nested lists?Sachin Rajput– Sachin Rajput2021-01-04 05:21:20 +00:00Commented Jan 4, 2021 at 5:21
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6 Answers
One way without using for loop:
len(list(filter(lambda x: x[0] == 3, arr))) Output:
2 
Comments
Try this sum with a list comprehension:
print(sum([i[0] == 3 for i in lst])) Output:
2 
1 Comment
BOBTHEBUILDER
that counts every value but I'm only looking for specific indexes. in this example the correct output would be 2.
from collections import Counter lst = [[3, 3, 5], [3, 2, 8], [2, 1, 3]] print (Counter(sublist[0] for sublist in lst)[3]) You can use the Counter function. Mention which position you want to look at in sublist, and which key (i.e. integer 3) you want to print the result for.
Output:
2 2 Comments
codeforester
Isn't this expensive since we are only interested in a fixed position in each of the elements in the list of lists?
Synthaze
Certainly, but who knows what the OP wants to make with it after. And if limited to the question, being expansive is not a problem considering the length of the data.
Using numpy only:
import numpy as np a = [[3, 3, 5], [3, 2, 8], [2, 1, 3]] b = a[:, 0] c = np.where(b == 3) print(c[0].shape[0]) Using pandas only:
import pandas as pd a = [[3, 3, 5], [3, 2, 8], [2, 1, 3]] df = pd.DataFrame(a) print(df[df[0] == 3].shape[0]) Output:
2 Comments
All what you need is to check the order of the entered number in terms of slicing and comparison.
According to python 3.8
Code Syntax
lists = [[3, 3, 5], [3, 2, 8], [2, 1, 3]] num = int(input("check our number: ")) counter = 0 for each in lists: if num == each[0]: counter +=1 print (f" your {num} occur found: {counter} times ") Output
check our number: 3 your 3 occur found: 2 times [Program finished] 
Comments
I think chain from itertools might be the solution if you don't want to use loops
from itertools import chain t=[[1, 2, 3], [3, 5, 6], [7], [8, 9]] new_list = list(chain(*t)) print(new_list) Output [1, 2, 3, 3, 5, 6, 7, 8, 9]
After this you can just use count function this 'new_list' to check the count of any elements
