Python 中没有重复的随机对numpy 或 itertools

问题描述

我有一个人的列表，我想生成一些随机的配对或三元组，以便列表中的每个人都只属于一对或三元组（不多也不少）。

示例：

@interface ViewController () {
    UIAlertController *tableAlert;
    UIAlertController *notesAlert;
}
@end

@implementation ViewController

- (void)viewDidLoad {
   [super viewDidLoad];
   self.tableAlert = [self initAlert];
   self.notesAlert = [self initAlert];
}

// func to init the alerts
- (UIAlertController *) initAlert {
    UIAlertController *alert = [UIAlertController alertControllerWithTitle: @"" message: @"" preferredStyle: UIAlertControllerStyleActionSheet];
    [alert setModalPresentationStyle:UIModalPresentationPopover];
    [alert.popoverPresentationController setSourceView: self.view];
    
    UIPopoverPresentationController *popover = [alert popoverPresentationController];
    CGRect popoverFrame = CGRectMake(0,self.view.frame.size.width/2,self.view.frame.size.width/2);
    popover.sourceRect = popoverFrame;
    UIAlertAction *dismiss = [UIAlertAction actionWithTitle: @"Ok" style: UIAlertActionStyleDefault handler:nil];
    [alert addAction: dismiss];

    return alert;
}

我使用 >>> people = ["John","Paul","George","Ringo","David","Roger","Richard","Nick","Syd"] >>> generate_random_pairs(people) [ ("John","George"),("Paul","David"),("Roger","Nick"),("Ringo","Syd") ] 、numpy.random.choice 或 numpy.random.sample 尝试了不同的想法，但似乎都没有奏效。

解决方法

基于 jonrsharpe 的评论

import numpy as np

people = ["John","Paul","George","Ringo","David","Roger","Richard","Nick","Syd"]

shuffled = np.random.permutation(people)

pairs_or_triplets = [list(a) for a in np.array_split(shuffled,len(shuffled) / 2)]

,

使用 grouper() recipe from itertools：

from itertools import zip_longest
from random import shuffle

def grouper(iterable,n,fillvalue=None):
    "Collect data into fixed-length chunks or blocks of size n."
    # grouper('ABCDEFG',3,'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args,fillvalue=fillvalue)

people = ["John","Syd"]

shuffle(people)

list(grouper(people,''))

示例输出：

[('Ringo','George','Syd'),('Richard','David','John'),('Paul','Roger','Nick')]