Is there a pythonic way for loop and use previous item

Question

For example, I have a array X [1, 3, 2, 5, 1, 3], I need a new array like a moving window, find the smallest in every 3 items ==> [1, 1, 1, 2, 1, 1]

I know I can loop from array len() and use list slip, but Is there a pythonic way for this problem?

I tried X[1] and found type(X[1]) is a int and cannot trace back the the array X.

Thank you for your help.

edited: sorry for the inconvenient. i made a mistake above. for window 1, only the first element 1, so output 1 window 2: [1,3] -> 1 window 3: [1, 3, 2] -> 1 window 4: [3, 2, 5] -> 2

btw the anwsers is very helpful. thank you.

Could you please explain in more detail how you get your expected output? It seems to me that there should only be _one_ leading `1`. — inspectorG4dget, May 10 '14 at 14:16
If I have understood your question, the expected output should be `[1, 2, 1, 1]`, not `[1, 1, 2, 2, 1, 1]`... — Stefano Sanfilippo, May 10 '14 at 14:17
@sshashank124 you can write a program that will produce any given output. The OP was not clear in specifying what he/she meant, but certainly you cannot say it does not make sense now. — Stefano Sanfilippo, May 10 '14 at 14:27
Thank you so much, I made a mistake in OP. and the answer for rolling window is very helpful — football, May 12 '14 at 04:57

Liteye · Answer 1 · 2014-05-10T14:35:19.473

1

map(min, zip(a[1:]+[max(a)], [max(a)]+a, a+[max(a)]))

I don't know if it is a pythonic way. It's just a tricky one-liner and is not preferable to ordinary loop.

edited May 10 '14 at 14:35

answered May 10 '14 at 14:17

Liteye

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+1 coz it gives desired output – vaultah May 10 '14 at 14:36
-1 coz it's ugly and inefficient – vaultah May 10 '14 at 14:37

score 1 · Accepted Answer · answered May 10 '14 at 14:18

1

In [33]: X = [1, 3, 2, 5, 1, 3]

In [34]: list(map(min, (X[i:i+3] for i in range(len(X)-2))))
Out[34]: [1, 2, 1, 1]

answered May 10 '14 at 14:18

inspectorG4dget

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This does not yield the expected output in the question. Maybe the OP wanted something different, you should wait untile he/she clarifies. – Stefano Sanfilippo May 10 '14 at 14:22
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@StefanoSanfilippo: I already commented, asking for clarification from OP. Since OP seems unresponsive to comments on the question itself, I figured I'd give it a shot, hoping to hit something close to what OP is looking for – inspectorG4dget May 10 '14 at 14:28
@inspectorG4dget Thanks a lot, that's what I need. and I hope the window rolls from the first and single element, I'll solve this. – football May 12 '14 at 04:58
@StefanoSanfilippo Thank you, I made a mistake in OP. the third number should be 1. I've updated it. – football May 12 '14 at 05:05

score 0 · Answer 3 · answered May 10 '14 at 14:26

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def pairwise( iterable, n=2 ):
    from itertools import tee, izip, islice
    return izip(*(islice(it,pos,None) for pos,it in enumerate(tee(iterable, n))))

 x = [1, 3, 2, 5, 1, 3]


list(pairwise(x,3))
#[(1, 3, 2), (3, 2, 5), (2, 5, 1), (5, 1, 3)]
map(min, pairwise(x,3))
#[1, 2, 1, 1]

answered May 10 '14 at 14:26

M4rtini

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Thanks, itertools is very helpful. learning. – football May 12 '14 at 04:59

score 0 · Answer 4 · answered May 10 '14 at 14:53

history = collections.deque((None, None), 2)  # sentinel
for x in your_input:
    result = min(x, min(history))
    history.append(x)

Choice of sentinel depends on how you want to treat start of input.

Or if this is clearer

window = collections.deque((), 3)
for x in your_input:
    window.append(x)
    result = min(window)

First uses explicit history as OP subject, last explicit window.

deque with size argument automatically drops off elements from the other end on append when size is exceeded.

Is there a pythonic way for loop and use previous item

4 Answers4