Combining multiple for loops that iterate over matrix python

Question

hello I want to combine multiple for loops into one loop to make my code less complex. the loop iterates over the elements in a list of lists and add +1 for every correct value.

so let's say i have

ink = 0
mouse = 0
rat = 0
matrix = [
['', 'article1.txt', 'article2.txt', 'article3.txt'], ['ink', 2, 3, 0], ['mouse', 0, 2, 1], ['rat', 0, 0, 1]]

and

for line in matrix[1:2]: 
    for i in line[1:]:
            if i > 0:
                ink = ink + 1
for line in matrix[2:3]: 
    for i in line[1:]:
            if i > 0:
                mouse = mouse + 1
for line in matrix[3:4]: 
    for i in line[1:]:
            if i > 0:
                rat = rat + 1

I'd want this to become one loop or atleast some shorter code that automatically does this, even if there would be more rows and columns in the matrix.

for row in matrix[1:]: for col in row[1:]: ...

Iron Fist
– Iron Fist

2021-02-06 11:13:37 +00:00
Commented Feb 6, 2021 at 11:13 — Iron Fist
– Iron Fist, Commented Feb 6, 2021 at 11:13

Honn · Accepted Answer · 2021-02-06 11:17:26Z

3

So assuming a correct value is a value greater than 1, i did this: Also to have infinite values, i organized the results in a dictionary:

matrix = [
    ["", "article1.txt", "article2.txt", "article3.txt"],
    ["ink", 2, 3, 0],
    ["mouse", 0, 2, 1],
    ["rat", 0, 0, 1]]
results = {product[0]: [1 for a in product[1:] if a > 0].count(1) for product in matrix[1:]}

print(results)

I hope this is what you expected. Let me know if you need any adjustments or explanations. This would be the result for your example:

{'ink': 2, 'mouse': 2, 'rat': 1}

answered Feb 6, 2021 at 11:17

Honn

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1 Comment

Kareltje12 Over a year ago

Yes this works and I am able to understand! Thanks

Iron Fist · Accepted Answer · 2021-02-06 11:40:48Z

3

One way to improve your code, would be:

>>> d = {}
>>> for row in matrix[1:]:
        d[row[0]] = len(list(filter(lambda x: x>0, row[1:])))

                 
>>> d
                 
{'ink': 2, 'mouse': 2, 'rat': 1}

This also could be compacted into one liner dictionary comprehension:

>>> d = {row[0]:len(list(filter(lambda x: x>0, row[1:]))) for row in matrix[1:]}

edited Feb 6, 2021 at 11:40

answered Feb 6, 2021 at 11:31

Iron Fist

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3 Comments

DarrylG Over a year ago

Great so upvoted, but also could have d as a dict comprehension i.e.: d = {row[0]:len(list(filter(lambda x: x>0, row[1:]))) for row in matrix[1:]}

Iron Fist Over a year ago

@DarrylG, true, I just wanted to stick to OP code to show him how he can improve his, but I'll add your comment in the answer as well

DarrylG Over a year ago

@IronFist--understood. Sort of an interesting code golf problem with even shorter solutions such as d = {row[0]:sum(x>0 for x in row[1:]) for row in matrix[1:]}

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