itertools

问题 I need to cycle through a list for starting position between 1-4 using itertools I am able to cycle through the list positions = itertools.cycle([1,2,3,4]) next(positions) This does return the next position, but what if the next time I need to start at 3? How can I set the start position? I need the start position to change often, I cant just change the list to start at 3. 回答1: You can't set a starting position; it'll always start where the given sequence starts. You can move the cycle along

问题 I need to cycle through a list for starting position between 1-4 using itertools I am able to cycle through the list positions = itertools.cycle([1,2,3,4]) next(positions) This does return the next position, but what if the next time I need to start at 3? How can I set the start position? I need the start position to change often, I cant just change the list to start at 3. 回答1: You can't set a starting position; it'll always start where the given sequence starts. You can move the cycle along

问题 I need to cycle through a list for starting position between 1-4 using itertools I am able to cycle through the list positions = itertools.cycle([1,2,3,4]) next(positions) This does return the next position, but what if the next time I need to start at 3? How can I set the start position? I need the start position to change often, I cant just change the list to start at 3. 回答1: You can't set a starting position; it'll always start where the given sequence starts. You can move the cycle along

问题 Is there something like Python's itertools.product() that provides the iteration through the Cartesian product of a set of sets in Gray code order ? For example, supposing that such a hypothetical generator existed, and it was called gray_code_product() , then gray_code_product(['a','b','c'], [0,1], ['x','y']) would generate, in the order : ('a',0,'x') ('a',0,'y') ('a',1,'y') ('a',1,'x') ('b',1,'x') ('b',1,'y') ('b',0,'y') ('b',0,'x') ('c',0,'x') ('c',0,'y') ('c',1,'y') ('c',1,'x') 回答1:

问题 I am using itertools to run a numerical simulation iterating over all possible combinations of my input parameters. In the example below, I have two parameters and six possible combinations: import itertools x = [0, 1] y = [100, 200, 300] myprod = itertools.product(x, y) for p in myprod: print p[0], p[1] # run myfunction using p[0] as the value of x and p[1] as the value of y How can I get the size of myprod (six, in the example)? I'd need to print this before the for loop starts. I

问题 I am using itertools to run a numerical simulation iterating over all possible combinations of my input parameters. In the example below, I have two parameters and six possible combinations: import itertools x = [0, 1] y = [100, 200, 300] myprod = itertools.product(x, y) for p in myprod: print p[0], p[1] # run myfunction using p[0] as the value of x and p[1] as the value of y How can I get the size of myprod (six, in the example)? I'd need to print this before the for loop starts. I

问题 Given a list, I can get the product of each item in a list as such: from itertools import product x = 'apple orange pair None'.split() [i + ' ' + j for i, j in product(x, x)] [out]: ['apple apple', 'apple orange', 'apple pair', 'apple None', 'orange apple', 'orange orange', 'orange pair', 'orange None', 'pair apple', 'pair orange', 'pair pair', 'pair None', 'None apple', 'None orange', 'None pair', 'None None'] If I want to nest the output of product(list, list) with the initial list, I could

问题 I am using a list of list with varying sizes. For example alternativesList can include 4 lists in one iteration and 7 lists in the other. What i am trying to do is capture every combination of words in different lists. Lets say that a= [1,2,3] alternativesList.append(a) b = ["a","b","c"] alternativesList.append(b) productList = itertools.product(*alternativesList) will create [(1, 'a'), (1, 'b'), (1, 'c'), (2, 'a'), (2, 'b'), (2, 'c'), (3, 'a'), (3, 'b'), (3, 'c')] One problem here is that my

问题 Note : I know how I can do this of course in an explicit for loop but I am looking for a solution that is a bit more readable. If possible, I'd like to solve this by using some of the built-in functionalities. Best case scenario is something like result = [ *groupby logic* ] Assuming the following list: import numpy as np np.random.seed(42) N = 10 my_tuples = list(zip(np.random.choice(list('ABC'), size=N), np.random.choice(range(100), size=N))) where my_tuples is [('C', 74), ('A', 74), ('C',

问题 So I've generated a list of combinations, using itertools and I'm getting a result that looks like this: nums = [-5,5,4,-3,0,0,4,-2] x = [x for x in set(itertools.combinations(nums, 4)) if sum(x)==target] >>> x = [(-5, 5, 0, 4), (-5, 5, 4, 0), (5, 4, -3, -2), (5, -3, 4, -2)] What is the most time-complexity wise efficient way of removing unordered duplicates, such as x[0] and x[1] are the duplicates. Is there anything built in to handle this? My general approach would be to create a counter