repeat

问题 What is an efficient way to repeat a string to a certain length? Eg: repeat(\'abc\', 7) -> \'abcabca\' Here is my current code: def repeat(string, length): cur, old = 1, string while len(string) < length: string += old[cur-1] cur = (cur+1)%len(old) return string Is there a better (more pythonic) way to do this? Maybe using list comprehension? 回答1: def repeat_to_length(string_to_expand, length): return (string_to_expand * ((length/len(string_to_expand))+1))[:length] For python3: def repeat_to

问题 import numpy as np data = np.arange(-50,50,10) print data [-50 -40 -30 -20 -10 0 10 20 30 40] I want to repeat each element of data 5 times and make new array as follows: ans = [-50 -50 -50 -50 -50 -40 -40 ... 40] How can I do it? What about repeating the whole array 5 times? ans = [-50 -40 -30 -20 -10 0 10 20 30 40 -50 -40 -30 -20 -10 0 10 20 30 40 -50 -40 -30 -20 -10 0 10 20 30 40 -50 -40 -30 -20 -10 0 10 20 30 40 -50 -40 -30 -20 -10 0 10 20 30 40 .......] 回答1: In [1]: data = np.arange(-50

问题 This question already has an answer here: Repeat rows of a data.frame N times 7 answers I want to repeat the rows of a data.frame, each N times. The result should be a new data.frame (with nrow(new.df) == nrow(old.df) * N ) keeping the data types of the columns. Example for N = 2: A B C A B C 1 j i 100 1 j i 100 --> 2 j i 100 2 K P 101 3 K P 101 4 K P 101 So, each row is repeated 2 times and characters remain characters, factors remain factors, numerics remain numerics, ... My first attempt

问题 In Python, where [2] is a list, the following code gives this output: [2] * 5 # Outputs: [2,2,2,2,2] Does there exist an easy way to do this with an array in JavaScript? I wrote the following function to do it, but is there something shorter or better? var repeatelem = function(elem, n){ // returns an array with element elem repeated n times. var arr = []; for (var i = 0; i <= n; i++) { arr = arr.concat(elem); }; return arr; }; 回答1: You can do it like this: function fillArray(value, len) { if

问题 Let\'s say I have a one-dimensional array: a = [1, 2, 3]; Is there a built-in Matlab function that takes an array and an integer n and replicates each element of the array n times? For example calling replicate(a, 3) should return [1,1,1,2,2,2,3,3,3] . Note that this is not at all the same as repmat . I can certainly implement replicate by doing repmat on each element and concatenating the result, but I am wondering if there is a built in function that is more efficient. 回答1: As of R2015a ,

问题 I would expect this line of JavaScript: \"foo bar baz\".match(/^(\\s*\\w+)+$/) to return something like: [\"foo bar baz\", \"foo\", \" bar\", \" baz\"] but instead it returns only the last captured match: [\"foo bar baz\", \" baz\"] Is there a way to get all the captured matches? 回答1: When you repeat a capturing group, in most flavors, only the last capture is kept; any previous capture is overwritten. In some flavor, e.g. .NET, you can get all intermediate captures, but this is not the case

问题 I\'m trying to insert multiple values into an array using a \'values\' array and a \'counter\' array. For example, if: a=[1,3,2,5] b=[2,2,1,3] I want the output of some function c=somefunction(a,b) to be c=[1,1,3,3,2,5,5,5] Where a(1) recurs b(1) number of times, a(2) recurs b(2) times, etc... Is there a built-in function in MATLAB that does this? I\'d like to avoid using a for loop if possible. I\'ve tried variations of \'repmat()\' and \'kron()\' to no avail. This is basically Run-length

问题 I\'ve seen regex patterns that use explicitly numbered repetition instead of ? , * and + , i.e.: Explicit Shorthand (something){0,1} (something)? (something){1} (something) (something){0,} (something)* (something){1,} (something)+ The questions are: Are these two forms identical? What if you add possessive/reluctant modifiers? If they are identical, which one is more idiomatic? More readable? Simply \"better\"? 回答1: To my knowledge they are identical. I think there maybe a few engines out