Question: I have some dictionary

```
someDict = {
    'foo1': [1, 4, 7, 0, -2],
    'foo2': [0, 2, 5, 3, 6],
    'foo3': [1, 2, 3, 4, 5]
}

```

I would like to loop over all the elements in each list with Python 3, and when the element at some given index equals zero, I want to delete that element at that index for *all* the lists/properties in the dictionary. So that the dictionary ends up as

```
someDict = {
    'foo1': [4, 7, -2],
    'foo2': [2, 5, 6],
    'foo3': [2, 3, 5]
}

```

Please note that I don't know beforehand how many keys/lists the dictionary will have, and I don't know how many elements the list will contain. I have come up with the following code, which seems to work, but was wondering if there is a more efficient way to do this?

```
keyPropList = someDict.items()
totalList = []

for tupleElement in keyPropList:
    totalList.append(tupleElement[1])

copyTotalList = totalList[:]
for outerRow in copyTotalList:
    for outerIndex, outerElement in enumerate(outerRow):
        if outerElement==0:
            for innerIndex, _ in enumerate(copyTotalList):
                del totalList[innerIndex][outerIndex]

print('someDict =', someDict)

```

Answer: 
You can find a list of "banned" indices which can then be used to filter the structure:

```
someDict = {
  'foo1': [1, 4, 7, 0, -2],
  'foo2': [0, 2, 5, 3, 6],
  'foo3': [1, 2, 3, 4, 5]
}
results = {i for c in someDict.values() for i, a in enumerate(c) if not a}
new_dict = {a:[c for i, c in enumerate(b) if i not in results] for a, b in someDict.items()}

```

Output:

```
{'foo1': [4, 7, -2], 'foo2': [2, 5, 6], 'foo3': [2, 3, 5]}

```
Other option, more verbose and destructive (`pop`) for `original_dict`:

```
some_dict = {
    'foo1': [1, 0, 2, 3, -2],
    'foo2': [0, 2, 5, 3, 6],
    'foo3': [1, 2, 3, 0, 5]
}

for k, v in some_dict.items():
  id = []
  for i, e in enumerate(v):
    if e == 0: id.append(i-len(id))
  [ [ some_dict[k].pop(x) for x in id ] for k,v in some_dict.items() ]

print(some_dict)
#=> {'foo1': [2, -2], 'foo2': [5, 6], 'foo3': [3, 5]}

```