itertools

问题 I have Set of tuples. For example: set([(('E', ('T',)), 0), (('F', ('(', 'E', ')')), 0), (('T', ('F',)), 0), (('__S__', ('E', '$')), 0), (('E', ('E', '+', 'T')), 0), (('T', ('T', '*', 'F')), 0), (('F', ('id',)), 0)]) as you see every tuple has a tuple as it's first element ( ex. ('F', ('(', 'E', ')')) ). first element of this tuple is single character and second element is another tuple ( ex. ('(', 'E', ')')) ). this tuple has one or more single character in it. (It is actually Context Free

问题 I try to use Pool from the multiprocessing module to speed up reading in large csv files. For this, I adapted an example (from py2k), but it seems like the csv.dictreader object has no length. Does it mean I can only iterate over it? Is there a way to chunk it still? These questions seemed relevant, but did not really answer my question: Number of lines in csv.DictReader, How to chunk a list in Python 3? My code tried to do this: source = open('/scratch/data.txt','r') def csv2nodes(r):

问题 This question already has an answer here : Which itertools generator doesn't skip any combinations? (1 answer) Closed 5 years ago . I have tried to use itertools to compute all combinations of a list ['a', 'b', 'c'] using combinations_with_replacement with repeating elements. The problem is in the fact that the indices seem to be used to distinguish the elements: Return r length subsequences of elements from the input iterable allowing individual elements to be repeated more than once.

问题 This question already has answers here : Combine Python Dictionary Permutations into List of Dictionaries (2 answers) Closed 4 years ago . I have dict that describes possible config values, e.g. {'a':[1,2], 'b':[3,4,5]} I want to generate list of all acceptable configs, e.g. [{'a':1, 'b':3}, {'a':1, 'b':4}, {'a':1, 'b':5}, {'a':2, 'b':3}, {'a':2, 'b':4}, {'a':1, 'b':5}] I've looked through the docs and SO and it certainly seems to involve itertools.product , but I can't get it without a

问题 I've made a post here, yet as I got no answer as per now I thought maybe to try it also here as I've found it relevant. I have the following code: import pandas as pd import numpy as np import itertools from pprint import pprint # Importing the data df=pd.read_csv('./GPr.csv', sep=',',header=None) data=df.values res = np.array([[i for i in row if i == i] for row in data.tolist()], dtype=object) # This function will make the subsets of a list def subsets(m,n): z = [] for i in m: z.append(list

问题 I work with strings, each having a dynamic amount of optional variables in parenthesis: (?please) tell me something (?please) Now I want to replace the variables with an empty string and get back all possible variations: tell me something (?please) (?please) tell me something tell me something The wanted function is supposed to handle multiple, different and an endless amount of variables. any help highly appreciated. 回答1: The problem with using the solution at String Replacement Combinations

问题 I managed to modify the recipe for roundrobin in https://docs.python.org/3.1/library/itertools.html to include a limit (stop when reaching X elements) - code below... Now - what I really want is "stop when reaching X elements but with no element duplication". Is it even possible? (because it is a generator...) def roundrobin(limit, *iterables): "roundrobin('ABC', 'D', 'EF') --> A D E B F C" # Recipe credited to George Sakkis pending = len(iterables) nexts = cycle(iter(it).next for it in

问题 What I want to achieve is to generate combinations in different processes to get them faster. In this post there'a a method to get the kth combination of a set of n numbers in groups of p numbers. Once I have the kth combination, I can manage to generate the rest of combinations in different processes (dividing the total number of combinations between processes so I know the combination each process has to begin with). The way I've done this has been with the same function that gets the kth

问题 I just found this instruction itertools.product(*[(0, 1)] * n) posted by PAG. Can someone explain how it works? I am trying to find a way of doing permutations without repetition of n tuples in 3 bags and I only can use itertools if I want. Thanks 回答1: [(0, 1)] is a list of a single tuple of the numbers 0 and 1 . [(0, 1)] * n duplicates the tuple inside of the list, so we get [(0, 1), (0, 1), ..., (0, 1), (0, 1)] Then, if we look at the itertools.product function, we want to pass in each of

问题 I'm trying to port some code that uses Python's iterator tools to Node, however I'm not seeing any analogs out there. I'm specifically looking for a replacement for itertools.combination() I've seen this one, but it's incomplete and out of date: node-intertools 回答1: I was not happy with the available solutions, so I wrote my own: https://www.npmjs.com/package/iter-tools 来源： https://stackoverflow.com/questions/24642185/is-there-an-equivalent-to-pythons-iterator-tools-for-node-js