Python For loop not incrementing

Question

clean_offset = len(malware)

tuple_clean = []
tuple_malware = []

for i in malware:
    tuple_malware.append([malware.index(i), 0])
    print(malware.index(i))
    print(tuple_malware)
for j in clean:
    tuple_clean.append([(clean_offset + clean.index(j)), 1])
    print(clean.index(j))
    print(tuple_clean)
    import pdb; pdb.set_trace()

training_data_size_mal = 0.8 * len(malware)
training_data_size_clean = 0.8 * len(clean)

i increments as normal and produces correct output however j remains at 0 for three loops and then jumps to 3. I don't understand this.

Answer 1

for a in something

a is what is contained in something, not the index

for example:

for n in [1, 10, 9, 3]:
    print(n)

gives

1
10
9
3

Answer 2

You either want

for i in range(len(malware))

or

for i, element in enumerate(malware)

at which point the i is the count and the element in the malware.index(i)

The last one is considered best practice when needing both the index and the element at that index in the loop.

Answer 3

There is a logical error on clean.index(j) . Array.index will return the first matched index in that array. So if there are some equal variables there will be some error

You can inspect with below code.

malware = [1,2,3,4,5,6,7,8,8,8,8,8,2]
clean = [1,2,3,4,4,4,4,4,4,2,4,4,4,4]

clean_offset = len(malware)

tuple_clean = []
tuple_malware = []

for i in malware:
    tuple_malware.append([malware.index(i), 0])
    print(malware.index(i))
    print(tuple_malware)
for j in clean:
    tuple_clean.append([(clean_offset + clean.index(j)), 1])
    print(clean.index(j))
    print(tuple_clean)

training_data_size_mal = 0.8 * len(malware)
training_data_size_clean = 0.8 * len(clean)

Answer 4

op has already figured the question, but in case anyone is wondering or needs a TL;DR of Barkin's comment, its just a small correction,

replace

for i in malware

for j in clean

with

for i in range(len(malware))

for j in range(len(clean))

and at the end remove the.index() function, and place i and j.