itertools

I am looking to group similar items in a list based on the first three characters in the string. For example: test = ['abc_1_2', 'abc_2_2', 'hij_1_1', 'xyz_1_2', 'xyz_2_2'] How can I group the above list items into groups based on the first grouping of letters (e.g. 'abc' )? The following is the intended output: output = {1: ('abc_1_2', 'abc_2_2'), 2: ('hij_1_1',), 3: ('xyz_1_2', 'xyz_2_2')} or output = [['abc_1_2', 'abc_2_2'], ['hij_1_1'], ['xyz_1_2', 'xyz_2_2']] I have tried using itertools.groupby to accomplish this without success: >>> import os, itertools >>> test = ['abc_1_2', 'abc_2_2',

I noticed that itertools does not (it seems to me) have a function capable of interleaving elements from several other iterable objects (as opposed to zipping them): def leaf(*args): return (it.next() for it in cycle(imap(chain,args))) tuple(leaf(['Johann', 'Sebastian', 'Bach'], repeat(' '))) => ('Johann', ' ', 'Sebastian', ' ', 'Bach', ' ') (Edit) The reason I ask is because I want to avoid unnecessary zip/flatten occurrences. Obviously, the definition of leaf is simple enough, but if there is a predefined function that does the same thing, I would prefer to use that, or a very clear

I am experiencing very strange issues while working with the data inside my function that gets called by pool.map. For example, the following code works as expected... import csv import multiprocessing import itertools from collections import deque cur_best = 0 d_sol = deque(maxlen=9) d_names = deque(maxlen=9) **import CSV Data1** def calculate(vals): #global cur_best sol = sum(int(x[2]) for x in vals) names = [x[0] for x in vals] print(", ".join(names) + " = " + str(sol)) def process(): pool = multiprocessing.Pool(processes=4) prod = itertools.product(([x[2], x[4], x[10]] for x in Data1))

I have a array like this, a = [3,2,5,7,4,5,6,3,8,4,5,7,8,9,5,7,8,4,9,7,6] and I want to make list of values that are lesser than 7 (look like following) b = [[3,2,5],[4,5,6,3],[4,5],[5],[4],[6]] So I used following method, >>> from itertools import takewhile >>> a = [3,2,5,7,4,5,6,3,8,4,5,7,8,9,5,7,8,4,9,7,6] >>>list(takewhile(lambda x: x < 7 , a)) [3, 2, 5] But I only get the first sequence. Can anyone help me to solve this problem ? Thank you. a = [3,2,5,7,4,5,6,3,8,4,5,7,8,9,5,7,8,4,9,7,6] from itertools import groupby [list(g) for k, g in groupby(a, lambda x:x<7) if k] Output: [[3, 2, 5],

I want to get started trying to develop a few simple applications with PyObjC. I installed PyObjC and the Xcode templates. I know that PyObjC itself works, since I've run this script successfully. When I tried to create a project from the Cocoa-Python Application template and ran it, I got this error: Traceback (most recent call last): File "main.py", line 10, in import objc File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages/PyObjC/objc/__init__.py", line 25, in from _convenience import * File "/opt/local/Library/Frameworks/Python.framework/Versions/2

Given a dictionary mapping variables to possible outcomes: { 'lblA' : [False, True], 'lblB' : [False, True], 'lblC' : [0,1,2] } I want to enumerate all possible dictionary outcomes: [ { 'lblA' : False , 'lblB' : False, 'lblC' : 0 }, { 'lblA' : True , 'lblB' : False, 'lblC' : 0 }, { 'lblA' : False , 'lblB' : True, 'lblC' : 0 }, { 'lblA' : True , 'lblB' : True, 'lblC' : 0 }, { 'lblA' : False , 'lblB' : False, 'lblC' : 1 }, { 'lblA' : True , 'lblB' : False, 'lblC' : 1 }, { 'lblA' : False , 'lblB' : True, 'lblC' : 1 }, { 'lblA' : True , 'lblB' : True, 'lblC' : 1 }, { 'lblA' : False , 'lblB' :

Using pythons itertools , I'd like to create an iterator over the outer product of all permutations of a bunch of lists. An explicit example: import itertools A = [1,2,3] B = [4,5] C = [6,7] for x in itertools.product(itertools.permutations(A),itertools.permutations(B),itertools.permutations(C)): print x While this works, I'd like to generalize it to an arbitrary list of lists. I tried: for x in itertools.product(map(itertools.permutations,[A,B,C])): print x but it did not do what I intended. The expected output is: ((1, 2, 3), (4, 5), (6, 7)) ((1, 2, 3), (4, 5), (7, 6)) ((1, 2, 3), (5, 4), (6