random

问题 My question: How can I create a list of random numbers with a given mean and standard deviation (sd) in Javascript? Example: I want to create a list with 5 random numbers in a range between 1 to 10. The resulting mean should be 5 and the standard deviation should be 2. What I've done so far: My idea was (http://jsfiddle.net/QASDG/3/): Make a function (named "createData") which creates 5 random numbers between 1 and 10 and push them into an "array" This function should also calculate the mean

问题 My question: How can I create a list of random numbers with a given mean and standard deviation (sd) in Javascript? Example: I want to create a list with 5 random numbers in a range between 1 to 10. The resulting mean should be 5 and the standard deviation should be 2. What I've done so far: My idea was (http://jsfiddle.net/QASDG/3/): Make a function (named "createData") which creates 5 random numbers between 1 and 10 and push them into an "array" This function should also calculate the mean

问题 I would like to shuffle output of find BUT with a fixed seed, so that every time I run the command I get the same output. Here's how I shuffle: find . -name '*.wav' | shuf The issue is that every time we have a new seed -> new order. My attempt to fix it: find . -name '*.wav' | shuf --random-source=<(echo 42) That works only on occasions (i.e. just a few cases, in a deterministic way). Most of the time it fails with: shuf: ‘/proc/self/fd/12’: end of file Same error is produced by e.g. seq 1

问题 I want to write a program that displays all the elements of a list in random order without repetition. It seems to me that it should work, but only prints those elements with repetition. import random tab = [] for i in range(1, 8): item = random.choice(["house", "word", "computer", "table", "cat", "enter", "space"]) if item not in tab: print(item) else: tab.append(item) continue 回答1: Instead of random.choice within the for loop, use random.shuffle here. This way, your list is guaranteed to be

问题 I want to write a program that displays all the elements of a list in random order without repetition. It seems to me that it should work, but only prints those elements with repetition. import random tab = [] for i in range(1, 8): item = random.choice(["house", "word", "computer", "table", "cat", "enter", "space"]) if item not in tab: print(item) else: tab.append(item) continue 回答1: Instead of random.choice within the for loop, use random.shuffle here. This way, your list is guaranteed to be

问题 I want to write a program that displays all the elements of a list in random order without repetition. It seems to me that it should work, but only prints those elements with repetition. import random tab = [] for i in range(1, 8): item = random.choice(["house", "word", "computer", "table", "cat", "enter", "space"]) if item not in tab: print(item) else: tab.append(item) continue 回答1: Instead of random.choice within the for loop, use random.shuffle here. This way, your list is guaranteed to be

问题 I want to write a program that displays all the elements of a list in random order without repetition. It seems to me that it should work, but only prints those elements with repetition. import random tab = [] for i in range(1, 8): item = random.choice(["house", "word", "computer", "table", "cat", "enter", "space"]) if item not in tab: print(item) else: tab.append(item) continue 回答1: Instead of random.choice within the for loop, use random.shuffle here. This way, your list is guaranteed to be

问题 How can I generate random numbers from 0 to 1000000? I already tried the code below, but it still gives me numbers from 0 to 32767 (RAND_MAX): #include <stdio.h> #include <stdlib.h> #include <time.h> int main(){ int i,x; srand(time(NULL)); for(i=0; i<10000; i++){ int x = rand() % 10000000 + 1; printf("%d\n",x); } return 0; } 回答1: [Edit] The initial answer was for 0 to 1,000,000. I now see it should be 0 to 10,000,000. As rand() will give an answer of at least 15 bits, call rand() multiple

问题 How can I generate random numbers from 0 to 1000000? I already tried the code below, but it still gives me numbers from 0 to 32767 (RAND_MAX): #include <stdio.h> #include <stdlib.h> #include <time.h> int main(){ int i,x; srand(time(NULL)); for(i=0; i<10000; i++){ int x = rand() % 10000000 + 1; printf("%d\n",x); } return 0; } 回答1: [Edit] The initial answer was for 0 to 1,000,000. I now see it should be 0 to 10,000,000. As rand() will give an answer of at least 15 bits, call rand() multiple

问题 Simple test code: pop = numpy.arange(20) rng = numpy.random.default_rng(1) rng.choice(pop,p=numpy.repeat(1/len(pop),len(pop))) # yields 10 rng = numpy.random.default_rng(1) rng.choice(pop) # yields 9 The numpy documentation says: The probabilities associated with each entry in a. If not given the sample assumes a uniform distribution over all entries in a. I don't know of any other way to create a uniform distribution, but numpy.repeat(1/len(pop),len(pop)) . Is numpy using something else? Why