Deleting elements from lists in a dictionary, Python-CodePudding

I have a dictionary with certain keys and lists of couples as elements. I want to perform the following operation: remove every element (couple) from all the lists for which its second element (or first, doesn't really matter) fulfills a certain condition. I tried with a very simple piece of code to do that, but I did not succeed and I in fact noticed a specific behavior: removing an element form a list let's the list's elements for loop skip the following element, for some reason (or at least this is what it looks like to me).

This is a very simple example:

# random dictionary
a = {'a': [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [1, 7]],
     'b': [[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6], [2, 7]]}

def f(d):
    
    # keys in the dicitonary
    for key in list(d):
        
        # elements inthe list
        for elem in d[key]:
            
            # if the second element of the couple
            # satisfies a certain criterion...
            if elem[1] >= 2:
                
                # remove that element (first occurrence) from the list
                d[key].remove(elem)
        
    return d

b = f(a)
print(b)

This is the output I get:

{'a': [[1, 1], [1, 3], [1, 5], [1, 7]], 'b': [[2, 1], [2, 3], [2, 5], [2, 7]]}

This is what I would like to have:

{'a': [[1, 1]], 'b': [[2, 1]]}

How do I perform the above described operation correctly?

Edit: I know there are some workarounds. My question is now about the skipping thing occurring in the for loop: why does it happen? Why does the code not work as I think it should?

CodePudding user response：

When you remove a list element, the indices are all adjusted to reflect the deletion. If you work from the end of the list to the beginnng you will not experience the skipping you see e.g.

'mylist = [1,23,3,4,5,6,7,8,9]
     r = range(len(mylist)-1,0,-1)
     for i in r:
        if mylist[i] % 2 :
            mylist.pop(i)

CodePudding user response：

You could iterate over the values (which are lists) and in a nested loop iterate over those lists again and if a certain value doesn't satisfy your condition, ignore it.

# random dictionary
a = {'a': [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [1, 7]],
     'b': [[2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [2, 6], [2, 7]]}

def f(d):
    # iterate over the keys
    for key in d.keys():
        value_list = d[key]
        # create a temporary list
        temp_list = []
        for elem in value_list:
            # if some element doesn't match the criteria, skip it
            if elem[1] >= 2:
                continue
            # if the condition is satisfied, add it to the temporary list
            temp_list.append(elem)
        # replace the key's value with the modified list
        d[key] = temp_list
    
    return d

b = f(a)
print(b)

CodePudding user response：

Q. Why does the skipping happen?

>>> numbers = list(range(1, 6))
>>> numbers
[1, 2, 3, 4, 5]
>>> for i in numbers:
...     print(f'{i=}')
i=1
i=2
i=3
i=4
i=5

The for loop above is essentially turned into this:

>>> it = iter(numbers)
>>> while True:
...     try:
...         i = next(it)
...     except StopIteration:
...         break
...     print(f'{i=}')
i=1
i=2
i=3
i=4
i=5

The first step is creating an iterator.

>>> it = iter(numbers)

You could say that it is "pointing at" the first element in the list:

[1, 2, 3, 4, 5]
 ^

Calling next() returns that element and "moves" the "pointer" up a step.

>>> next(it)
1

it is now pointing at the 2nd element

[1, 2, 3, 4, 5]
    ^

Let's modify numbers by removing the first element:

>>> numbers.pop(0)
1
>>> numbers
[2, 3, 4, 5]

We have not changed it which is still pointing at the 2nd element of numbers

[2, 3, 4, 5]
    ^

When we call next() again we will get 3 and as a result 2 has been "skipped".

>>> next(it)
3