P: n/a

If this is the list.
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
as we can see there are peaks in the list.that is 0,72,0 is a
group(triangle) with peak 72.then 0, 4, 9, 2, 0, 0 with peak
9 and 0, 42, 26, 0 with 42 and so on...
what I want is the left and right bound index of each bin(triangle).The
list could as big as possible.So some heurestic algorithm which could
first find max in the list and look for local maxima and minima and
group its adjcent bounds.Then find next max and group the bins and so
on.
so that we can get
[[0,2],[2,7],[7,10],[10,13]]
( indexes of the bounds in the values list). so first group [0,2]
correspond to 0,72,0 in the values list and so on...
Hope I am clear.  
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P: n/a
 qu*****@gmail.com wrote: If this is the list.
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282, 23, 0, 101, 0, 0, 0, 0, 0]
as we can see there are peaks in the list.that is 0,72,0 is a group(triangle) with peak 72.then 0, 4, 9, 2, 0, 0 with peak 9 and 0, 42, 26, 0 with 42 and so on... what I want is the left and right bound index of each bin(triangle).The list could as big as possible.So some heurestic algorithm which could first find max in the list and look for local maxima and minima and group its adjcent bounds.Then find next max and group the bins and so on.
so that we can get
[[0,2],[2,7],[7,10],[10,13]] ( indexes of the bounds in the values list). so first group [0,2] correspond to 0,72,0 in the values list and so on... Hope I am clear.
Not exactly your output, but hopefully you can tailor it to your needs:
py> values = [0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282, 23, 0, 101, 0, 0,
0, 0, 0]
py> def isnonzero((index, val)):
.... return val != 0
....
py> import itertools
py> for nonzero, vals in itertools.groupby(enumerate(values), isnonzero):
.... if nonzero:
.... vals = list(vals)
.... start = vals[0][0]  1
.... end = vals[1][0] + 1
.... print [start, end], "values: ", values[start:end+1]
....
[0, 2] values: [0, 72, 0]
[2, 6] values: [0, 4, 9, 2, 0]
[7, 10] values: [0, 42, 26, 0]
[10, 13] values: [0, 282, 23, 0]
[13, 15] values: [0, 101, 0]
Note that the real work is done by itertools.groupby. Zero terms are
grouped together and ignored; nonzero terms are gathered together in lists.
STeVe  
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Thanks for that. My version of python does'nt find "groupby". I am
using python 2.3.2. Is there a way I could do it with out using groupby  
P: n/a
 qu*****@gmail.com wrote: Thanks for that. My version of python does'nt find "groupby". I am using python 2.3.2. Is there a way I could do it with out using groupby
itertools.groupby is in Python 2.4. The docs[1] give a Python
equivalent, so if for some reason you can't upgrade to the current
version of Python, you can just copy the groupby code from there.
STeVe
[1]http://docs.python.org/lib/itertoolsfunctions.html#l2h1379  
P: n/a

what if we do something like this. Assume the values list is the
content of a histogram. Then we see that
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
1 is repeated 72 times, 3 > 4 times and so on. That is the index
would be the value repeated as many times as in the list.
Now If we find the max and look for the adjcent index. FOr example
take 282 :
Index of 282 > 11
now check for value of index 10 and 12 .If value(10) < value(9) then
there is a dip as value(9) and value(11) are greater than value(10).
that could be the lower bound of that range. then if
value(11) and value(13) for value(12). if value(11)< value(10) there is
a dip. SO the bounds of [10,12] > [0,282,23] rather than [10, 13]
values: [0, 282, 23, 0] ( in your case).
How would this work. Is there someway I could do this faster.  
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Hi Steve!
I am not sure if I was clear with my previous post .Ok let me rephrase
it .
Assume the values list is the
content of a histogram. Then we see that
values = [ 0, 72, 2, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
1 is repeated 72 times, 3 > 4 times and so on. That is the index
would be the value repeated as many times as in the list.
Now If we find the max and look for the adjcent index.
That is if we plot the above list as a histogram. We will have crests
and troughs ie peaks and dips. if we find two dips then the region
between the two dips could be a range like [0, 72, 2] .So here we
are not looking for a zero. But if we find dips then we consider the
regions between it as a bin and group it.
 /\
 /\ / \ /\
 / \/ \/ \
/_____________________________________

1 2 3
so pictorially. In the above plot. If y axis is the list above. then we
need to bin it this way.
If I use you previous approach using the groupby then all these three
regions will be considered as one.
Hope I am clear this time.  
P: n/a
 qu*****@gmail.com wrote: If this is the list.
values = [ 0, 72, 0, 4, 9, 2, 0, 0, 42, 26, 0, 282, 23, 0, 101, 0, 0, 0, 0, 0]
as we can see there are peaks in the list.that is 0,72,0 is a group(triangle) with peak 72.then 0, 4, 9, 2, 0, 0 with peak 9 and 0, 42, 26, 0 with 42 and so on... what I want is the left and right bound index of each bin(triangle).The list could as big as possible.So some heurestic algorithm which could first find max in the list and look for local maxima and minima and group its adjcent bounds.Then find next max and group the bins and so on.
so that we can get
[[0,2],[2,7],[7,10],[10,13]] ( indexes of the bounds in the values list). so first group [0,2] correspond to 0,72,0 in the values list and so on... Hope I am clear.
ISTM you just have to look for the valleys  places where the values change from descending to
ascending. Here is a simpleminded way to do it:
def findPeaks(values):
groups = []
startIx = 0
lastValue = values[0]
ascending = True
for i, value in enumerate(values[1:]):
if value <= lastValue:
ascending = False
else:
if not ascending:
# Switch from descending to ascending
groups.append( [startIx, i] )
startIx = i
ascending = True
lastValue = value
# Get the last group if any
if i > startIx:
groups.append( [startIx, i] )
return groups
values = [ 0, 72, 2, 4, 9, 2, 0, 0, 42, 26, 0, 282,
23, 0, 101, 0, 0, 0, 0, 0]
print findPeaks(values)
## prints:
[[0, 2], [2, 7], [7, 10], [10, 13], [13, 18]]
Kent  
P: n/a
 qu*****@gmail.com wrote: Hi Steve! I am not sure if I was clear with my previous post .Ok let me rephrase it .
Assume the values list is the content of a histogram. Then we see that values = [ 0, 72, 2, 4, 9, 2, 0, 0, 42, 26, 0, 282, 23, 0, 101, 0, 0, 0, 0, 0]
1 is repeated 72 times, 3 > 4 times and so on. That is the index would be the value repeated as many times as in the list. Now If we find the max and look for the adjcent index.
That is if we plot the above list as a histogram. We will have crests and troughs ie peaks and dips. if we find two dips then the region between the two dips could be a range like [0, 72, 2] .So here we are not looking for a zero. But if we find dips then we consider the regions between it as a bin and group it.
 /\  /\ / \ /\  / \/ \/ \ /_____________________________________  1 2 3
so pictorially. In the above plot. If y axis is the list above. then we need to bin it this way. If I use you previous approach using the groupby then all these three regions will be considered as one.
Hope I am clear this time.
So you want the peaks and valleys basically. Well, using itools.window
from http://aspn.activestate.com/ASPN/Coo.../Recipe/299529, you
could do something like:
py> values = [0,72,2,4,9,2,0,0,42,26,0,282,23,0,101,0,0,0,0,0]
py> peaks = [i
.... for i, (v1, v2, v3) in enumerate(itools.window(values))
.... if v1 < v2 and v2 > v3]
py> peaks
[1, 4, 8, 11, 14]
py> valleys = [i
.... for i, (v1, v2, v3) in enumerate(itools.window(values))
.... if v1 >= v2 and v2 <= v3]
py> valleys
[2, 6, 7, 10, 13, 15, 16, 17, 18]
py> [(min((abs(p  v), v) for v in valleys + [0] if v < p)[1],
.... p,
.... min((abs(p  v), v) for v in valleys if v > p)[1])
.... for p in peaks]
[(0, 1, 2), (2, 4, 6), (7, 8, 10), (10, 11, 13), (13, 14, 15)]
This is not likely the most efficient approach, but it's pretty simple:
* identify the peaks and valleys
* identify the valley on each side of a peak
STeVe  
P: n/a

Hi Kent,
Thanks for that. But We are considering [..., 0, 101, 0, 0, 0, 0, 0] >
[13,18] .In fact if you look at the list, the histogram ends at 15 that
is [0,101,0] > [13,15]. Dont you think so.  
P: n/a
 qu*****@gmail.com wrote: Hi Kent, Thanks for that. But We are considering [..., 0, 101, 0, 0, 0, 0, 0] > [13,18] .In fact if you look at the list, the histogram ends at 15 that is [0,101,0] > [13,15]. Dont you think so.
Well you consider ..., 0, 4, 9, 2, 0, 0, ... as an interval [2, 7] in your example. How is
that different from ..., 0, 101, 0, 0, 0, 0, 0 ?
Kent  
P: n/a

I get a syntax error in :
py> [(min((abs(p  v), v) for v in valleys + [0] if v < p)[1],
.... p,
.... min((abs(p  v), v) for v in valleys if v > p)[1])
.... for p in peaks]  
P: n/a

oh yes its the same case. even [0,4,9,2,0] as a set [2,6] and may be
not [2,7]. Its not that you are wrong its jus that I was not clear.
Sorry about that.  
P: n/a
 qu*****@gmail.com wrote: I get a syntax error in :
py> [(min((abs(p  v), v) for v in valleys + [0] if v < p)[1], ... p, ... min((abs(p  v), v) for v in valleys if v > p)[1]) ... for p in peaks]
I think we already covered the part where you were using an older
version of Python. In this case, the missing feature is "generator
expressions" and they are inside the two min() calls.
You might want to consider upgrading...
Peter  
P: n/a

Peter Hansen wrote: qu*****@gmail.com wrote:
I get a syntax error in :
py> [(min((abs(p  v), v) for v in valleys + [0] if v < p)[1], ... p, ... min((abs(p  v), v) for v in valleys if v > p)[1]) ... for p in peaks]
I think we already covered the part where you were using an older version of Python. In this case, the missing feature is "generator expressions" and they are inside the two min() calls.
You might want to consider upgrading...
But if you can't, you should write this as something like:
[(min([(abs(p  v), v) for v in valleys + [0] if v < p])[1],
p,
min([(abs(p  v), v) for v in valleys if v > p])[1])
for p in peaks]
Note the extra brackets in the min calls.
STeVe   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 3088
 replies: 13
 date asked: Jul 19 '05
