Python - Finding each occurrence of a value in a mixed array (integers, lists)

Question

I have an array: x = [ [1, 2], 1, 1, [2, 1, [1, 2]] ] in which I want to count every occurrence of the number 1, and store that number in the variable one_counter. x.count(1) returns only 2 occurrences of 1, which is insufficient.

My code below serves my purpose and stores 5 in one_counter, however it looks messy and feels unpythonic to me.

Any suggestions how I can improve its pythonicity and expand it into more-dimensional lists?

Thanks!

x = [[1, 2], 1, 1, [2, 1, [1, 2]]]

one_counter = 0

for i in x:
    if type(i) == list:
        for j in i:
            if type(j) == list:
                for k in j:
                    if k == 1:
                        one_counter += 1

            else:
                if j == 1:
                    one_counter += 1

    else:
        if i == 1:
            one_counter += 1

@IgnacioVazquez-Abrams I guess he overused the word Scale,he might have meant expansion of his datatype into more dimensional lists. — DhruvPathak
– DhruvPathak, Commented Oct 9, 2011 at 7:38
So how did you end up with this list, and why do you need this count? The whole thing seems oddly artificial. — Karl Knechtel
– Karl Knechtel, Commented Oct 9, 2011 at 8:50
@Karl Would you be upset if I said it was me practicing, hitting a road block, and asking for help? — Shon Freelen
– Shon Freelen, Commented Oct 11, 2011 at 3:53

Mark Byers · Accepted Answer · 2011-10-09 07:37:19Z

8

You could use recursion:

def flatten_count(iterable, element):
    count = 0
    for item in iterable:
        if item == element:
            count += 1
        if isinstance(item, list):
            count += flatten_count(item, element)
    return count

Or more concisely:

def flatten_count(iterable, element):
    return sum(
        flatten_count(item, element) if isinstance(item, list) else item == element
        for item in iterable 
    )

Use like this:

>>> x = [[1, 2], 1, 1, [2, 1, [1, 2]]]
>>> print(flatten_count(x, 1))
5

edited Oct 9, 2011 at 7:37

answered Oct 9, 2011 at 7:31

Mark Byers

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1 Comment

Shon Freelen Over a year ago

Thanks so much, Mark! That makes perfect sense, and now I understand recursion better along with the handiness that is isinstance().

DhruvPathak · Accepted Answer · 2011-10-09 07:37:01Z

3

A hacky solution, working by conversion of datatype to string : http://codepad.org/vNEv6B8M

import re
x = [ [1, 2], 1, 1, [2, 1, [1, 2]] ]
nums = [int(i) for i in re.findall(r'\d+', str(x))]
print(nums.count(1))

answered Oct 9, 2011 at 7:37

DhruvPathak

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Comments

ovgolovin · Accepted Answer · 2011-10-09 11:09:10Z

2

I think it's better to separate this task into 2 parts.

Part 1

Part 1 is to create an generator which will flatten the input list.

def flatten_list(L):
    for i in L:
        if isinstance(i,list):
            for j in flatten_list(i):
                yield j
        else:
            yield i

Testing the output:

x = [[1, 2], 1, 1, [2, 1, [1, 2]]]

for i in flatten_list(x):
    print i

Output:

Part 2

Part 2 is to use the flattened list to count the number of occurrences of 1 in it:

print(sum(i==1 for i in flatten_list(x)))

Output:

Note that i==1 returns True if i=1, and False if i is not equal to 1. But Trueis equal to 1 and False is equal to 0, so sum just calculates the number of True occurrences (which is equal to 5 in this case).

answered Oct 9, 2011 at 11:09

ovgolovin

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1 Comment

Shon Freelen Over a year ago

Thanks for breaking it down. I really appreciate it.

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