pairwise

问题 Here is my problem : let's say my two array are : import numpy as np first = np.array(["hello", "hello", "hellllo"]) second = np.array(["hlo", "halo", "alle"]) Now I want to get the matrix of distance between each element of the two arrays so for example my distance function is : def diff_len(string1, string2): return abs(len(string1) - len(string2)) So I I would like to get the matrix : hello hello hellllo hlo result1 result2 result3 halo result4 result5 result6 alle result7 result8 result9

来源： https://stackoverflow.com/questions/63964995/assistance-in-fixing-the-performance-time-of-algorithmn

来源： https://stackoverflow.com/questions/63964995/assistance-in-fixing-the-performance-time-of-algorithmn

问题 Given the target ('b', 'a') and the inputs: x0 = ('b', 'a', 'z', 'z') x1 = ('b', 'a', 'z', 'z') x2 = ('z', 'z', 'a', 'a') x3 = ('z', 'b', 'a', 'a') The aim is to find the location of the continuous ('b', 'a') element and get the output: >>> find_ba(x0) 0 >>> find_ba(x1) 0 >>> find_ba(x2) None >>> find_ba(x3) 1 Using the pairwise recipe: from itertools import tee def pairwise(iterable): "s -> (s0,s1), (s1,s2), (s2, s3), ..." a, b = tee(iterable) next(b, None) return zip(a, b) I could do this

问题 I have a table of around 50k distinct rows, and 2 columns. You can think of each row being a movie, and columns being the attributes of that movie - "ID": id of that movie, "Tags":some content tags of the movie, in form of a list of strings for each movie . Data looks something like this: movie_1, ['romantic','comedy','English']; movie_2, ['action','kongfu','Chinese'] My goal is to first calculate the jacquard similarity between each Movie based on their corresponding tags, and once that's

问题 I have a table of around 50k distinct rows, and 2 columns. You can think of each row being a movie, and columns being the attributes of that movie - "ID": id of that movie, "Tags":some content tags of the movie, in form of a list of strings for each movie . Data looks something like this: movie_1, ['romantic','comedy','English']; movie_2, ['action','kongfu','Chinese'] My goal is to first calculate the jacquard similarity between each Movie based on their corresponding tags, and once that's

问题 I would like to get all comparison with pairwise.t.test even comparison with the same group to perform multcompLetter from the package multcompview . Because (I think), multcompLetters needs a full matrix and not only the half or a part of the comparison matrix to work. dflong <- structure(list(moda = structure(c(4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 4L, 1L

问题 I'm trying to use ddply (a plyr function) to sort and identify the most frequent interaction type between any unique pairs of user from a social media data of the following form from <- c('A', 'A', 'A', 'B', 'B', 'B', 'B', 'C', 'C', 'C', 'C', 'D', 'D', 'D', 'D') to <- c('B', 'B', 'D', 'A', 'C', 'C', 'D', 'A', 'D', 'B', 'A', 'B', 'B', 'A', 'C') interaction_type <- c('like', 'comment', 'share', 'like', 'like', 'like', 'comment', 'like', 'like', 'share', 'like', 'comment', 'like', 'share', 'like

问题 I have a 1D vector having N dimension in TensorFlow, how to construct sum of a pairwise squared difference? Example Input Vector [1,2,3] Output 6 Computed As (1-2)^2+(1-3)^2+(2-3)^2. if I have input as an N-dim vector l, the output should be sigma_{i,j}((l_i-l_j)^2). Added question : if I have a 2d matrix and want to perform the same process for each row of the matrix, and then average the results from all the rows, how can I do it? Many thanks! 回答1: For pair-wise difference, subtract the

问题 I have a big character matrix (15000 x 150), and with the following format: A B C D [1,] "0/0" "0/1" "0/0" "1/1" [2,] "1/1" "1/1" "0/1" "0/1" [3,] "1/2" "0/3" "1/1" "2/2" [4,] "0/0" "0/0" "2/2" "0/0" [5,] "0/0" "0/0" "0/0" "0/0" I need to do pairwise comparison between columns and get the proportion of rows where neither string separated by '/' is equal (coded as 0); only one string separated by '/' is equal (coded as 1); both strings separated by '/' are equal (coded as 2). The expected