Python - Categorizing the list of array part 2

Question

I hoping someone can help me with a problem I'm stuck(again) with.

if i have coordinates:

 x = array[0,1,2,3,4,3,2,3,-2,-4,-7,2,2]
 y = array[0,1,2,3,4,5,6,7,8,9,10,11,12]

and with help in Categorizing the list of array in python, i could make:

x = [(0, 1, 2, 3, 4), (4, 3, 2), (2, 3), (3, -2, -4, -7), (-7, 2), (2, 2)]

The question is, how should i made y like this:

y = [(0,1,2,3,4),(4,5,6),(6,7),(7,8,9,10),(10,11),(11,12)]

since, x and y is actually coordinates and they bounded each other.

i have tried to use loop function, and i realize the code is still wrong

se = []
for i in range(len(z)):
        k = z[i]
        for i in range(len(k)):
                se.append(y[i])

best regards,

Glenn

You should really make this question all inclusive. I had no idea what you were talking about until I went and read through your previous question. — jdi
– jdi, Commented May 17, 2012 at 20:12

Community · Accepted Answer · 2017-05-23 12:04:53Z

3

I am referencing the answer by @jamylak from your previous question, and showing a slight modification.

While you could try and match the pattern of your resulting x to your y, you could also just modify the original solution to treat x and y as points (x,y):

from itertools import groupby

x = [0,1,2,3,4,3,2,3,-2,-4,-7,2,2]
y = [0,1,2,3,4,5,6,7,8,9,10,11,12]

def slope(a,b): #Returns 1 for inc, 0 for equal and -1 for dec
    return (a > b) - (a < b) 

def groups(nums):
    # 
    # Change the call to slope() to assume 2d point tuples as values
    #
    for k,v in groupby(zip(nums,nums[1:]), lambda (x,y): slope(x[0],y[0])):
        yield next(v) + tuple(y for x,y in v) 

#
# Pass in a zipped data structure
#
print list(groups(zip(x,y)))
# result
[((0, 0), (1, 1), (2, 2), (3, 3), (4, 4)),
 ((4, 4), (3, 5), (2, 6)),
 ((2, 6), (3, 7)),
 ((3, 7), (-2, 8), (-4, 9), (-7, 10)),
 ((-7, 10), (2, 11)),
 ((2, 11), (2, 12))]

Though I am not sure if the resulting formatting is desirable to you.

Here is how you can separate them:

from operator import itemgetter

result = list(groups(zip(x,y)))
x = [map(itemgetter(0), points) for points in result]
y = [map(itemgetter(1), points) for points in result]
print x
# [[0, 1, 2, 3, 4], [4, 3, 2], [2, 3], [3, -2, -4, -7], [-7, 2], [2, 2]]
print y
# [[0, 1, 2, 3, 4], [4, 5, 6], [6, 7], [7, 8, 9, 10], [10, 11], [11, 12]]

Or as suggested by @jamylak:

x,y = zip(*[zip(*points) for points in result])

And to illustrate what @jamylak was talking about, regarding how the modification to the groups() method allows for N-dimension points or data sets:

z = ['foo',1,2,'bar',4,5,6,'foo',8,9,10,'bar',12]
print list(groups(zip(x,y,z)))
# result
[((0, 0, 'foo'), (1, 1, 1), (2, 2, 2), (3, 3, 'bar'), (4, 4, 4)),
 ((4, 4, 4), (3, 5, 5), (2, 6, 6)),
 ((2, 6, 6), (3, 7, 'foo')),
 ((3, 7, 'foo'), (-2, 8, 8), (-4, 9, 9), (-7, 10, 10)),
 ((-7, 10, 10), (2, 11, 'bar')),
 ((2, 11, 'bar'), (2, 12, 12))]

You can see that it can be any arbitrary data set, and its always just grouping on the first element of each.

edited May 23, 2017 at 12:04

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answered May 17, 2012 at 20:28

jdi

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3 Comments

glwilliam Over a year ago

thanks for your reply. it's true the result is not desirable. since i'm hoping 'x' and 'y' is separated and they in same format like above. x = [(0, 1, 2, 3, 4), (4, 3, 2), (2, 3), (3, -2, -4, -7), (-7, 2), (2, 2)] y = [(0,1,2,3,4),(4,5,6),(6,7),(7,8,9,10),(10,11),(11,12)] because later on, i'll input them in my Lagrange formula. but i'll try to separate it now.

jamylak Over a year ago

+1 I like this solution, since it uses the 0th element of each two consecutive points it supports any number of arrays eg. this also works with a list of z points. I would suggest replacing your last one liner with this: x,y = zip(*[zip(*points) for points in groups(zip(x,y))])

jdi Over a year ago

@jamylak: Great suggestions! I updated to illustrate your point.

Andrew Clark · Accepted Answer · 2012-05-17 20:38:04Z

2

The following does what you want:

x = [(0, 1, 2, 3, 4), (4, 3, 2), (2, 3), (3, -2, -4, -7), (-7, 2), (2, 2)]
y = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

s = map(len, x)
s = [0] + [sum(s[:i])-i for i in range(1, len(s))] + [len(y)]
y = [tuple(y[a:b+1]) for a, b in zip(s, s[1:])]

Result:

>>> y
[(0, 1, 2, 3, 4), (4, 5, 6), (6, 7), (7, 8, 9, 10), (10, 11), (11, 12)]

This basically constructs a list of positions where y will be split. We figure this out using the length of each tuple in x, but it gets a little tricky because the last element of a set is included as the next element in the following set.

Here is one of the intermediate values which may help to clarify how this works:

>>> zip(s, s[1:])
[(0, 4), (4, 6), (6, 7), (7, 10), (10, 11), (11, 13)]

We use this to construct the new y like this:

[(0, 4), (4,  6), (6,  7), (7, 10), (10, 11), (11, 13)]
   \ |     \   \    \_  \
   | |      \  |      \  |
[y[0:4+1], y[4:6+1], y[6:7+1], ...]

edited May 17, 2012 at 20:38

answered May 17, 2012 at 20:28

Andrew Clark

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1 Comment

glwilliam Over a year ago

i seeee! that's how it works! thanks man! thanks also for your explanation

Scott Hunter · Accepted Answer · 2012-05-17 20:27:54Z

0

A little ugly, but it works:

se = []
y2 = [y[0]]
i = 1
for xrun in x:
    first = True
    for xv in xrun:
        if first:
            first = False
            continue
        y2.insert( len(y2), y[i] )
        i += 1
    se.insert( len(se), tuple(y2) )
    y2 = [y[i-1]]

answered May 17, 2012 at 20:27

Scott Hunter

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Comments

Jay M · Accepted Answer · 2012-05-17 22:31:37Z

0

How about this one using numpy, it solves your first problem at the same time.

import numpy as np

x=(0, 1, 2, 3, 4, 3, 2, 3, -2, -4, -7, 2, 2)
y=range(13)

#First order differential, find slopes
dx = list((np.diff(x)>0)*1)

#First order differental looses first value, but we always want to keep it
#just decide if it's lower or higher than the 2nd value

d0=((x[0]-x[1])>0)*1

#Add the first order differential to the 2nd order differential (peaks)
ddx = [d0,]+list(np.abs(np.diff(dx)))

p=0
rx=[]
ry=[]

for n,v in enumerate(ddx):
    if v==1:
        rx.append(tuple(x[p:n+1]))
        ry.append(tuple(y[p:n+1]))
        p=n

print rx
print ry

answered May 17, 2012 at 22:31

Jay M

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