itertools

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed 22 days ago . So I need to figure out all the possible permutations to divide 31 1's and 713 0's over 744 positions (all zeroes and ones are exactly the same). I created the following code to create all permutations of 0's and 1's over 744 positions and my idea was to delete those lines which do not sum to 31,

问题 I'm look for a way to find all combinations of sums with elements of a Fibonacci sequence with a given limit which equal that same value. I know that combinations() from itertools is our best bet in solving such problems, but as I'm new in Python I like to know how I can retain the matching combinations (as only one is correct, as this is not the end of the algorithm). I currently have: # Call Fibonacci sequence with a given limit: def fib1(n): result = [] a, b = 1, 1 while a < n: result

问题 I recently had to debug some code that went something like this: for key, group in itertools.groupby(csvGrid, lambda x: x[0]): value1 = sum(row[1] for row in group) value2 = sum(row[2] for row in group) results.append([key, value1, value2]) In every result set, value2 came out as 0 . When I looked into it, I found that the first time the code iterated over group , it consumed it, so that the second time there were zero elements to iterate over. Intuitively, I would expect group to be a list

问题 In python, I have a dictionary like this... pleio = {'firstLine': {'enf1': ['54', 'set'], 'enf2': ['48', 'free'], 'enf3': ['34', 'set'], 'enf4': ['12', 'free']} 'secondLine':{'enf5': ['56','bgb'] 'enf6': ['67','kiol'] 'enf7': ['11','dewd'] 'enf8': ['464','cona']}} I would like to make paired combinations with no repetition of the elements in the inner dictionary, to end up with a result like this... {'enf3': ['34', 'set'], 'enf2': ['48', 'free']} {'enf3': ['34', 'set'], 'enf1': ['54', 'set']}

问题 I am trying to create a list of numbers from 0-9999 using itertools.product . I am able to create a list from 0000-9999 by doing the following: numbers = ['0','1','2','3','4','5','6','7','8','9'] itertools.product(numbers,numbers,numbers,numbers) And while I want entries like 0001 , I would also like to get 001 , 01 , and 1 . What would be the most effective way to include these? Should I make calls to itertools.product(numbers,numbers,numbers) and itertools.product(numbers,numbers) and then

问题 Is the itertools C module included somehow in the main Python binary in 3.x? Assuming that the C module is built and included, which it appears to be: >>> import inspect >>> import itertools >>> >>> inspect.getsourcefile(itertools) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/local/Cellar/python3/3.4.3_2/Frameworks/Python.framework/Versions/3.4/lib/python3.4/inspect.py", line 571, in getsourcefile filename = getfile(object) File "/usr/local/Cellar/python3

问题 I would like to use itertools.chain for efficient concatenation of lists (memoization), but I need to be able to read (or map , etc.) the result multiple times. This example illustrates the problem: import itertools a = itertools.chain([1, 2], [3, 4]) print list(a) # => [1, 2, 3, 4] print list(a) # => [] What is the best way to avoid this problem? 回答1: As with all generators, you'll need to convert it to a list and store that result instead: a = list(a) This is a fundamental principle of

问题 I can't seem to find an elegant way to start from t and result in s . >>>t = ['a',2,'b',3,'c',4] #magic >>>print s {'a': 2, 'c': 4, 'b': 3} Solutions I've come up with that seems less than elegant : s = dict() for i in xrange(0, len(t),2): s[t[i]]=t[i+1] # or something fancy with slices that I haven't figured out yet It's obviously easily solved, but, again, it seems like there's a better way. Is there? 回答1: I'd use itertools , but, if you think that's complicated (as you've hinted in a

问题 How can I skip the tuples which has duplicate elements in the iteration when I use itertools.product? Or let's say, is there anyway not to look at them in the iteration? Because skipping may be time consuming if the number of lists are too much. Example, lis1 = [1,2] lis2 = [2,4] lis3 = [5,6] [i for i in product(lis1,lis2,lis3)] should be [(1,2,5), (1,2,6), (1,4,5), (1,4,6), (2,4,5), (2,4,6)] It will not have (2,2,5) and (2,2,6) since 2 is duplicate in here. How can I do that? 回答1: itertools

问题 While answering the question Clunky calculation of differences between an incrementing set of numbers, is there a more beautiful way?, I came up with two solutions, one with List Comprehension and other using itertools.starmap. To me, list comprehension Syntax looks more lucid, readable, less verbose and more Pythonic. But still as starmap is well available in itertools, I was wondering, there has to be a reason for it. My Question is when starmap could be preferred over List Comprehension ?