Python find Index of elements in List and remove in another List

Question

I'm using Python 2.7 and couldn't find an answer to my question.

I have 3 lists (x and y coordinantes and the radius of detected circles):

XSum = [1316, 1324, 1330, 1344, 1356, 1344, 1356, 1308, 1322, 1310, 1336, 1934]
YSum = [960,  952,  972,  954,  964,  970,  948,  928,  940,  950,  984,  1080]
RSum = [379,  375,  355,  352,  340,  347,  338,  348,  361,  342,  342,  286]

I have to find elements in XSum which are not between 675 and 1350 (Index 4, 6 and 11). Elements with the index have to be removed from all three lists.

XSum = [1316, 1324, 1330, 1344, **1356**, 1344, **1356**, 1308, 1322, 1310, 1336, **1934**]
YSum = [960,  952,  972,  954, **964**,  970,  **948**,  928,  940,  950,  984,  **1080**]
RSum = [379,  375,  355,  352,  **340**,  347,  **338**,  348,  361,  342,  342,  **286**]

After first removal of index which are out of range:

XSum2 = [1316, 1324, 1330, 1344, 1344, 1308, 1322, 1310, 1336]
YSum2 = [960,  952,  972,  954,  970,  928,  940,  950,  984]
RSum2 = [379,  375,  355,  352,  347,  348,  361,  342,  342]

The next step is to find elements in YSum which are not between 940 and 980 and removed the detected index from all three lists (Index 9).

XSum2 = [1316, 1324, 1330, 1344, 1344, 1308, 1322, 1310, **1336**]
YSum2 = [960,  952,  972,  954,  970,  928,  940,  950,  **984**]
RSum2 = [379,  375,  355,  352,  347,  348,  361,  342,  **342**]

The resulting list should look like:

XSum_Result = [1316, 1324, 1330, 1344, 1344, 1308, 1322, 1310]
YSum_Result = [960,  952,  972,  954,  970,  928,  940,  950]
RSum_Result = [379,  375,  355,  352,  347,  348,  361,  342]

I think it should be an easy task, but I just couldn't figure it out how to remove by index from the other lists.

thank you all for the help

This is why it's generally better to make a single list of tuples or dictionaries, rather than having separate lists of each attribute. — Barmar
– Barmar, Commented Mar 24, 2017 at 15:51
You can do this type of filtering with a couple list comprehensions. — pylang
– pylang, Commented Mar 24, 2017 at 15:55
Umm... you know you're not dropping Y with a value of 928 there, right? — Jon Clements
– Jon Clements, Commented Mar 24, 2017 at 16:10

Jon Clements · Accepted Answer · 2017-03-24 16:09:54Z

3

You can group them together using zip and perform your logic to subset the candidates, then zip again (which inverses the original zip) and unpack into lists, eg:

x, y, r = (list(el) for el in zip(*((a, b, c) for a, b, c in zip(XSum, YSum, RSum) if 675 <= a <= 1350 and 940 <= b <= 980)))

answered Mar 24, 2017 at 16:09

Jon Clements

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4 Comments

pylang Over a year ago

This is good. You can even substitute the conditional expression with range itself, e.g. if a in range(675, 1351) and b in range(940, 981)

Jon Clements Over a year ago

@pylang that's okay in Python 3 but on Python 2 it will build a list and run a slow O(N) containership test

pylang Over a year ago

Agreed. On another note, compared to the expected results, 1308, 928, 348 are missing from x, y, r respectively.

Jon Clements Over a year ago

@pylang I think that's the OP's example output gone astray as 928 for Y doesn't meet the at least 940 criteria

Alex C. · Accepted Answer · 2017-03-24 16:06:26Z

1

You should have post a piece or code to help us know how to help you.

Here is a very basic idea, assuming that all list are sorted to match each other (I mean in index order) :

a = [1, 2, 3]
b = ['a', 'b', 'c']
for i in range(0, len(a)):
    if a[i] == 2:
        b.pop(i)
        break
print(b)

It should print out

['a', 'b']

Now, you can imagine this mechanism nested into a class holding an array of arrays well managed by an insert and a pop out methods and it becomes as easy as abc.

answered Mar 24, 2017 at 16:06

Alex C.

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