Counting backwards

Question

I've got a list organized like this :

[('down', 0.0098000000000000309), 
('up', 0.0015000000000000568), 
('down', 0.008900000000000019), 
('down', 0.023300000000000098), 
('down', 0.011599999999999944), 
('down', 0.0027000000000000357), 
('up', 0.0023999999999999577), 
('up', 0.0065000000000000613), 
('down', 0.0057000000000000384), 
('down', 0.018400000000000083), 
('up', 0.009300000000000086), 
('down', 0.0038000000000000256), 
('down', 0.00050000000000005596), 
('up', 0.0082000000000000961), .....

What would be the best way to "compare backwards?" , basically I want to return "yes" ( or whatever .. ) IF we`ve got a series of 2 "downs" followed by one "up" AND the second value is inferior to 0.0095 .

I hope he makes sense ..

[This question](http://stackoverflow.com/questions/323750/) shows how to iterate over a list using a sliding window. The rest should be simple enough. — Björn Pollex, May 26 '11 at 09:58

score 5 · Answer 1 · answered May 26 '11 at 10:10

5

Create a sliding window, and test on that:

def slidingwindow(iterable):
    iterator = iter(iterable)
    first, second = iterator.next(), iterator.next()
    for next in iterator:
        yield (first, second, next)
        first, second = second, next

def testforcondition(data):
    for window in slidingwindow(data):
        direction = [w[0] for w in window]
        if direction == ['down', 'down', 'up'] and window[2][1] < 0.0095:
            return True
    return False

answered May 26 '11 at 10:10

Martijn Pieters

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But much more readable! :-) Even if the slidingwindow function lived in a utility module, I'd be able to suss out what the testforcondition function was doing, one year from now, in an instant. – Martijn Pieters May 26 '11 at 10:36
Just goes to show, readability is often a subjective quality. – sje397 May 26 '11 at 10:38
Maintainability is subjective too, I guess. – Martijn Pieters May 26 '11 at 10:39

score 3 · Accepted Answer · answered May 26 '11 at 10:04

3

Here you go:

def frob(l):
    downcount = 0
    for ele in l:
        if downcount >= 2 and ele[0] == 'up' and ele[1] < 0.0095:
                return True
        downcount = (downcount + 1) if ele[0] == 'down' else 0
    return False

answered May 26 '11 at 10:04

verdesmarald

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sje397 · Answer 3 · 2011-05-26T10:13:27.560

0

Here's my attempt:

def test(data):
  for x in xrange(2, len(data)):
    if data[x-2][0] is 'down' and data[x][x-1] is 'down' and data[x][0] is 'up' and data[x][1] < 0.0095:
      return True
  return False

edited May 26 '11 at 10:13

answered May 26 '11 at 10:02

sje397

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Where is the test for 'up' after two 'down' values? – Martijn Pieters May 26 '11 at 10:12

Tim Pietzcker · Answer 4 · 2011-05-26T10:25:47.633

0

My suggestion (although now with the third slice, this is not really pretty anymore):

def compback(l):
    return any(i1[0] == i2[0] == "down" 
               and i2[1] < 0.0095
               and i3[0] == "up"
               for i1, i2, i3 in zip(l, l[1:], l[2:]))

edited May 26 '11 at 10:25

answered May 26 '11 at 10:03

Tim Pietzcker

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Same here: Where is the test for 'up' after two 'down' values? – Martijn Pieters May 26 '11 at 10:12
@Martijn Pieters: I completely overlooked that. I have now edited my answer accordingly. – Tim Pietzcker May 26 '11 at 10:26

score 0 · Answer 5 · answered May 26 '11 at 10:12

0

for index in xrange(0, len(list) - 2):
    if list[index][0] == 'down' and list[index + 1][0] == 'down' and list[index + 2][0] == 'up' and list[index + 1][1] < 0.0095:
         return True

answered May 26 '11 at 10:12

Sebastian Blask

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Counting backwards

5 Answers5