Question

I have a coordinated storage list in python A[row,col,value] for storing non-zeros values.

How can I get the list of all the row indexes? I expected this A[0:][0] to work as print A[0:] prints the whole list but print A[0:][0] only prints A[0].

The reason I ask is for efficient calculation of the number of non-zero values in each row i.e iterating over range(0,n) where n is the total number of rows. This should be much cheaper than my current way of for i in range(0,n): for j in A: ....

Something like:

c = []
# for the total number of rows
for i in range(0,n):
     # get number of rows with only one entry in coordinate storage list
     if A[0:][0].count(i) == 1: c.append(i)                
return c

Over:

c = []
# for the total number of rows 
for i in range(0,n):
    # get the index and initialize the count to 0 
    c.append([i,0])
    # for every entry in coordinate storage list 
    for j in A:
        # if row index (A[:][0]) is equal to current row i, increment count  
        if j[0] == i:
           c[i][1]+=1
return c

EDIT:

Using Junuxx’s answer, this question and this post I came up with the following (for returning the number of singleton rows) which is much faster for my current problems size of A than my original attempt. However it still grows with the number of rows and columns. I wonder if it’s possible to not have to iterate over A but just upto n?

# get total list of row indexes from coordinate storage list
row_indexes = [i[0] for i in A]
# create dictionary {index:count}
c = Counter(row_indexes)    
# return only value where count == 1 
return [c[0] for c in c.items() if c[1] == 1]

Answer 1

This should do it:

c = [x[0] for x in A]

It’s a list comprehension that takes the first (sub-)element of every element of A.