CODEWITHSUNDEEP
python
Finger twist
Finger twist

Reputation: 3786

Counting backwards

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 ..

Upvotes: 1

Views: 431

Answers (5)

Tim Pietzcker
Tim Pietzcker

Reputation: 336408

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:]))

Upvotes: 0

sje397
sje397

Reputation: 41862

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

Upvotes: 0

Sebastian Blask
Sebastian Blask

Reputation: 2938

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

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1123550

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

Upvotes: 5

verdesmarald
verdesmarald

Reputation: 11866

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

Upvotes: 3

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