Sure thing. It is a bit tricky to do this with LINQ but certainly possible using only the standard query operators. UPDATE: This is the subject of my blog on Monday June 28th 2010; thanks for the great question. Also, a commenter on my blog noted that there is an even more elegant query than … Read more
You need recursion: Let’s say all your lists are in lists, which is a list of lists. Let result be the list of your required permutations. You could implement it like this: void generatePermutations(List<List<Character>> lists, List<String> result, int depth, String current) { if (depth == lists.size()) { result.add(current); return; } for (int i = 0; … Read more
itertools.product should do the trick. >>> import itertools >>> list(itertools.product([1, 5, 8], [0.5, 4])) [(1, 0.5), (1, 4), (5, 0.5), (5, 4), (8, 0.5), (8, 4)] Note that itertools.product returns an iterator, so you don’t need to convert it into a list if you are only going to iterate over it once. eg. for x … Read more
This is very easy with list comprehensions. To get the cartesian product of the lists xs and ys, we just need to take the tuple (x,y) for each element x in xs and each element y in ys. This gives us the following list comprehension: cartProd xs ys = [(x,y) | x <- xs, y … Read more
You can use np.mgrid for this, it’s often more convenient than np.meshgrid because it creates the arrays in one step: import numpy as np X,Y = np.mgrid[-5:5.1:0.5, -5:5.1:0.5] For linspace-like functionality, replace the step (i.e. 0.5) with a complex number whose magnitude specifies the number of points you want in the series. Using this syntax, … Read more
You are looking for the Cartesian Product. In mathematics, a Cartesian product (or product set) is the direct product of two sets. In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there: import itertools x = [1, 2, 3, 4, 5, … Read more
Check out product() in the itertools module. It does exactly what you describe. import itertools my_list = [1,2,3,4] for pair in itertools.product(my_list, repeat=2): foo(*pair) This is equivalent to: my_list = [1,2,3,4] for x in my_list: for y in my_list: foo(x, y) Edit: There are two very similar functions as well, permutations() and combinations(). To illustrate … Read more
Use itertools.product, which has been available since Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) This is the same as: for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)
In recent versions of Pandas (>= 1.2) this is built into merge so you can do: from pandas import DataFrame df1 = DataFrame({‘col1′:[1,2],’col2’:[3,4]}) df2 = DataFrame({‘col3′:[5,6]}) df1.merge(df2, how=’cross’) This is equivalent to the previous pandas < 1.2 answer but is easier to read. For pandas < 1.2: If you have a key that is repeated … Read more
2020 Update: 1-line (!) answer with vanilla JS Original 2017 Answer: 2-line answer with vanilla JS: (see updates below) All of the answers here are overly complicated, most of them take 20 lines of code or even more. This example uses just two lines of vanilla JavaScript, no lodash, underscore or other libraries: let f … Read more