CODEWITHSUNDEEP
pythonloops
ismail
ismail

Reputation: 3923

Optimal way to access a value from the last iteration in a loop

What's the best and fastest way to access a value from the previous iteration in a for loop, assuming that the object will be very large (example, a cursor object which has 100,000+ records)

Using a simple example:

tmp = [
         ['xyz', 335], ['zzz', 338], ['yyy', 339], ['yyy', 442], 
         ['abc', 443], ['efg', 444], ['ttt', 446], ['fff', 447]
      ]

for x in tmp:
   if not prev:
     prev = x[1]
   print 'seq: ', x[1], 'prev seq:', prev, 'variance: ', x[1]-prev
   prev = x[1]

Is this the most optimal way to handle this?

Based on the responses below i did some testing: tmp was created with 500 lists, the average of running it 20 times is shown below.

results:

Mines: 0,623
Dave snippet1: 0,605
Dave snippet2: 0,586
Catchmeifyoutry (edited code): 0,707

Upvotes: 3

Views: 485

Answers (6)

jfs
jfs

Reputation: 414795

it = imap(operator.itemgetter(1), tmp) # get all 2nd items
prev = next(it, None) # get 1st element (doesn't throw exception for empty `tmp`)
for x in it:
    print 'seq: %s prev seq: %s variance: %s' % (x, prev, x-prev)
    prev = x

Upvotes: 0

tzot
tzot

Reputation: 96061

Guido's time machine to the rescue!

From the itertools recipes page:

import itertools
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)

This should be the most appropriate method (consider the iterable was (random.randint(100) for x in xrange(1000)); here iter(iterable); next(iterable) as a secondary iterator might not provide correct functionality.

Use it in your loop as:

for prev_item, item in pairwise(iterable):
    …

Upvotes: 1

Xavier Martinez-Hidalgo
Xavier Martinez-Hidalgo

Reputation: 618

Using itertools:

from itertools import izip, islice
for prev, cur in izip(l, islice(l, 1, None)):
    print 'seq:', cur[1], 'prev seq:', prev[1], 'delta:', cur[1]-prev[1]

For the specific example given in the question, note that, if the numbers can be represented using 32-bit ints, and the list of numbers fits into memory, one of the fastest ways to compute the difference would be to use numpy:

import numpy
a = numpy.array([x[1] for x in tmp])
delta = numpy.diff(a)

Upvotes: 2

catchmeifyoutry
catchmeifyoutry

Reputation: 7389

Just iterate over pairs, using zip(), which is much more readable.

UPDATE: for python 2.x, use itertools.izip instead as it is more efficient!

from itertools import izip
for prev, next in izip(tmp, tmp[1:]):
    print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1]

which can also use value unpacking to avoid the index:

for (_, prev), (_, next) in izip(tmp, tmp[1:]):
    print 'seq: ', next, 'prev seq:', prev, 'variance: ', next-prev

Or, if you really need the first iteration too

for prev, next in izip(tmp, tmp[:1] + tmp):
    print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1]

EDIT

If you want to avoid the creation of a list in the second argument you can also use an explicit iterator:

itr = iter(tmp)
itr.next() # here I assume tmp is not empty, otherwise an exception will be thrown
for prev, next in izip(tmp, itr):
    print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1]

Note: This zip pattern is useful in similar problems too. For example to extract successive triplets from a list:

xs = range(9)
triplets = zip(xs[::3], xs[1::3], xs[2::3]) # python 2.x, zip returns a list

print xs       # [0, 1, 2, 3, 4, 5, 6, 7, 8]
print triplets # [(0, 1, 2), (3, 4, 5), (6, 7, 8)]

Also note that in python 3 zip returns an iterator, similar to itertools.izip.

Upvotes: 4

culebrón
culebrón

Reputation: 36513

This code generates NameError because at if not prev, prev is not defined. Set it to False or None before the cycle. Also you may make a different loop:

for i in xrange(1, len(tmp)):
    print 'seq: {0}, prev seq: {1}, variance: {2}'.format(tmp[i][1], tmp[i - 1][1], tmp[i] - tmp[i - 1][1])

If you'll use 100,000+ records, the bottleneck will be not the cycle, but the memory used by the app. Don't store all the data in such a format: each pair of values (a list) will eat 100+ bytes. If they're in a file, it's better to iterate over it's lines:

(assuming the data is tab-separated)

def reader(filename):
    with open(filename) as f:
        prev = f.next()
        for l in f:
            l = l.split('\t')
            yield (prev, l)
            prev = l

for (prev, curr) in reader(myfile):
    print 'seq: {0}, prev seq: {1}, variance: {2}'.format(curr[1], prev[1], curr[1] - prev[1])

reader is a generator, it returns values from a sequence many times. This way, only 2 lines of data will be stored in memory at any moment, and your app will sustain even millions of rows.

To make the code readable, I put it aside, so that in the program body we dealed with the sequence of data, without caring how it's composed.

Upvotes: 0

Dave Kirby
Dave Kirby

Reputation: 26582

Your code is going to be doing the "if not prev" test every time round the loop, even though it only applies to the first element. Also your code seems broken to me - the first time round the loop the prev and current values are the same.

I would do it like this, assuming that there is at least one element:

tmp_iter = iter(tmp)
prev = tmp_iter.next()

for x in tmp_iter: 
   print 'seq: ', x[1], 'prev seq:', prev[1], 'variance: ', x[1]-prev[1]
   prev = x

this could be optimised further by getting rid of the indexing:

tmp_iter = iter(tmp)
[_, prev] = tmp_iter.next()

for [_, x] in tmp_iter: 
   print 'seq: ', x, 'prev seq:', prev, 'variance: ', x-prev
   prev = x

I use the assignment to spit the list into its constituent parts, and assign the first element to _ because it is not used.

Upvotes: 3

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