Oh! One other option: since you’re looking for close correspondences between two sorted lists, you could go through them both simultaneously, using a merge-like algorithm. This should be O(max(length(xm), length(xn)))-ish. match_for_xn = zeros(length(xn), 1); last_M = 1; for N = 1:length(xn) % search through M until we find a match. for M = last_M:length(xm) dist_to_curr … Read more
Extend A and B to 3D keeping their first axis aligned and introducing new axes along the third and second ones respectively with None/np.newaxis and then multiply with each other. This would allow broadcasting to come into play for a vectorized solution. Thus, an implementation would be – A[:,:,None]*B[:,None,:] We could shorten it a bit … Read more
Here’s one approach that does most of the processing on NumPy before finally putting it out as a DataFrame, like so – m,n,r = a.shape out_arr = np.column_stack((np.repeat(np.arange(m),n),a.reshape(m*n,-1))) out_df = pd.DataFrame(out_arr) If you precisely know that the number of columns would be 2, such that we would have b and c as the last two … Read more
The indexin function does something similar to what you want: indexin(a, b) Returns a vector containing the highest index in b for each value in a that is a member of b. The output vector contains 0 wherever a is not a member of b. Since you want a boolean for each element in your … Read more
You need to extend the dimensions of X with None/np.newaxis to form a 3D array and then do subtraction by w. This would bring in broadcasting into play for this 3D operation and result in an output with a shape of (5,n,3). The implementation would look like this – X[:,None] – w # or X[:,np.newaxis] … Read more
outer(0:5, 0:6, sum) don’t work because sum is not “vectorized” (in the sense of returning a vector of the same length as its two arguments). This example should explain the difference: sum(1:2,2:3) 8 1:2 + 2:3 [1] 3 5 You can vectorize sum using mapply for example: identical(outer(0:5, 0:6, function(x,y)mapply(sum,x,y)), outer(0:5, 0:6,’+’)) TRUE PS: Generally … Read more
The answers given are all correct. I just wanted to elaborate on gnovice’s remark about floating-point testing. When comparing floating-point numbers for equality, it is necessary to use a tolerance value. Two types of tolerance comparisons are commonly used: absolute tolerance and relative tolerance. (source) An absolute tolerance comparison of a and b looks like: … Read more
One simple method would be to assign the default value first and then perform 2 loc calls: In [66]: df = pd.DataFrame({‘x’:[0,-3,5,-1,1]}) df Out[66]: x 0 0 1 -3 2 5 3 -1 4 1 In [69]: df[‘y’] = 0 df.loc[df[‘x’] < -2, ‘y’] = 1 df.loc[df[‘x’] > 2, ‘y’] = -1 df Out[69]: x … Read more
Vectorization (as the term is normally used) refers to SIMD (single instruction, multiple data) operation. That means, in essence, that one instruction carries out the same operation on a number of operands in parallel. For example, to multiply a vector of size N by a scalar, let’s call M the number of operands that size … Read more
Your sample output is not legal. A matrix cannot have rows of different length. What you can do is create a cell array using arrayfun: values = arrayfun(@colon, idx1, idx2, ‘Uniform’, false) To convert the resulting cell array into a vector, you can use cell2mat: values = cell2mat(values); Alternatively, if all vectors in the resulting … Read more