Given a large list of fluctuating values, how do you determine all local min values? Not using numpy. Local minimum means all values in a list that are the troughs of a function.
List_y = [23, 8, -7, 57, 87, 6]
I would like:
New_list = [-7, 6]
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vaultah
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John Gubba
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If the number is less than the two on either side. – Malik Brahimi Apr 13 '15 at 14:38
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possible duplicate of [Finding local maxima/minima with Numpy in a 1D numpy array](http://stackoverflow.com/questions/4624970/finding-local-maxima-minima-with-numpy-in-a-1d-numpy-array) – TheBlackCat Apr 13 '15 at 14:50
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There's no numpy mention here. Don't assume that people automatically have external libraries. – Zizouz212 Apr 13 '15 at 14:52
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1You'll need to be more specific on what "local min values" means. Edit your question and gives us some expected input / output. – Hooked Apr 13 '15 at 16:00
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3What should `[23, 6, 6, 6, 42]` return? – Bill Lynch Apr 13 '15 at 19:49
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I didn't even think of that... wow! I guess it would need to be [6] because i only need one value of minima appended to another list. – John Gubba Apr 13 '15 at 20:07
2 Answers
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def local_min(ys):
return [y for i, y in enumerate(ys)
if ((i == 0) or (ys[i - 1] >= y))
and ((i == len(ys) - 1) or (y < ys[i+1]))]
>>> local_min([23, 8, -7, 57, 87, 6])
[-7, 6]
>>> local_min([23, 6, 6, 6, 42])
[6]
>>> local_min([6, 6, 4])
[4]
enrico.bacis
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@BillLynch: I changed it the other way, can you help me to spot other flaws? – enrico.bacis Apr 13 '15 at 20:28
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But `6` isn't a local minima at all in the list `[4, 6, 6]`. Oh well. – Bill Lynch Apr 13 '15 at 20:30
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@JohnGubba: Bill is right, 6 is more like a maximum. You should have better defined your question. – enrico.bacis Apr 13 '15 at 20:31
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@enrico.bacis: Personally, this problem is infinitely easier if we preprocess the list to remove adjacent duplicates. All of my annoying cases are depending on that. – Bill Lynch Apr 13 '15 at 20:32
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I'm a big fan of iterating over these problems in stages.
l = [23, 8, -7, -7, 57, 87, 6]
# Remove identical neighbors
# l becomes [23, 8, -7, 57, 87, 6]
l = [x for x,y in zip(l[0:], l[1:]) if x != y] + [l[-1]]
# Append positive infinity to both endpoints
# l becomes [inf, 23, 8, -7, 57, 87, 6, inf]
l = [float("inf")] + l + [float("inf")]
# Retain elements where each of their neighbors are greater than them.
# l becomes [-7, 6]
l = [y for x, y, z in zip(l[0:], l[1:], l[2:]) if x > y < z]
Bill Lynch
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As I understand it from its definition, the minima isn't strictly _smaller_ than its neighbors, but may be equal to (either of) them. Thus, I think your _Remove identical neighbors_ step may be eliminated, and the code be: `>>> l = [23, 8, -7, -7, 57, 87, 6]` >>> l = [float("inf")] + l + [float("inf")] `>>> l = [y for x, y, z in zip(l[0:], l[1:], l[2:]) if x >= y <= z]` >>> l `[-7, -7, 6]` – boardrider Apr 14 '15 at 09:31
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@boardrider: Then `3 4 4 4 3` would report `3 4 3` which is wrong. In addition, in the comments, John said that he was fine with only getting `-7, -6` from the input you suggest. I agree that a more accurate solution would have `-7 -7 6`, but creating that answer is harder with list comprehensions. – Bill Lynch Apr 14 '15 at 12:08
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@Bill, why is `3 4 3` wrong (if we take the definition of minimal to be <=) ? `3` is the minimal of `None and 3`, `4` is the minimal of `4 4 4`, and `3` is the minimal of `3 and None`. What don't I get? – boardrider Apr 14 '15 at 22:15
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@boardrider: John never fully clarified what he meant by minima. It's certainly possible that the definition you're suggesting is what he was looking for. Personally, I'm considering this more like a function. So we can't just look at neighbors. – Bill Lynch Apr 14 '15 at 22:35