Question

我将此列表作为示例

favourite_candy_boys_girls = [['Chips' , 1, 2], ['Chocolate', 3, 4], ['Lollipops', 2, 4]]

第二栏是男生，第三栏是女生。我想写一个函数，我返回第一个女孩最喜欢的糖果（在这种情况下它应该是巧克力）。我知道如果我运行以下内容：

max(l[2] for l in favourite_candy_boys_girls)

我会找到最大值，但我不太确定如何返回糖果

EDIT 我正在考虑使用如下循环将其编写为函数：

>>> def get_candy_with_max_girls(Candy: SystemData):
    """Return the candy type that has the most girls.
     If there is a tie of girls for a particular candy type, return the 
     candy type that appears first in candy.

>>> get_candy_with_max_girls(favourite_candy_boys_girls):
'Chocolate'
 """

Answer 1

您可以使用max函数的key参数。它允许您在排序之前提供将在您的项目上使用的功能：

l = [['Chips' , 1, 2], ['Chocolate', 3, 4], ['Lollipops', 2, 4]]
max(l, key=lambda x: x[2])[0]  
# returns 'Chocolate'

Answer 2

如果您使用字典，而密钥是食品，则可以使用以下代码：

<?php

$html = '
<div class="xGh" style="background-image: url(\'name_file0.jpg\');"></div>
<div class="xGh" style="background-image: url(\'name_file1.jpg\');"></div>
<div class="xGh" style="background-image: url(\'name_file2.jpg\');"></div>
<div class="xGh" style="background-image: url(\'name_file3.jpg\');"></div>
';

libxml_use_internal_errors(true);
$dom = new DOMDocument('1.0', 'utf-8');
$dom->loadHTML($html);
$dom->preserveWhiteSpace = false;
$xpath = new DOMXPath($dom);

$divs = $xpath->query('//div[@class=\'xGh\']/@style');

/* @var DOMAttr $div */
foreach ($divs as $div) {
    preg_match("/url\('(.*)'\)/", $div->nodeValue, $matches);

    if (!isset($matches[1])) {
        continue;
    }
    $url = $matches[1];

    // name_file0.jpg, ...
    echo $url . "\n";
}

哪个输出：

testlist = {"Chips":[1,2],"Chocolate":[3,4]}

output = max(testlist, key=lambda x: testlist[x][-1])
print output

Answer 3

试试这个（不需要图书馆，适合女孩和男孩，也适用于多种糖果价值）：

favourite_candy_boys_girls = [['Chips', 1, 2], ['Chocolate', 3, 4], ['Lollipops', 2, 4], ["gumdrop", 3, 4]]


my_dict = {}

# Convert Your List to Dictionary
for candy_box in favourite_candy_boys_girls:
    my_dict[candy_box[0]] = {"Girls": candy_box[1], "Boys": candy_box[2]}


def the_most_delicious_candy(get_dict, girl_or_boy):

    if girl_or_boy == "girls":

        just_girls = {}
        for candy, values in get_dict.items():
            just_girls[candy] = values["Girls"]

        find_max = max(just_girls.values())

        last_buffer = []
        for candy, values in get_dict.items():

            if values['Girls'] == find_max:

                make_new_dict = { candy: get_dict[candy] }
                last_buffer.append(make_new_dict)

        return last_buffer

    elif girl_or_boy == "boys":

        just_boys = {}
        for candy, values in get_dict.items():
            just_boys[candy] = values["Boys"]

        find_max = max(just_boys.values())

        last_buffer = []
        for candy, values in get_dict.items():

            if values['Boys'] == find_max:
                make_new_dict = {candy: get_dict[candy]}
                last_buffer.append(make_new_dict)

        return last_buffer

    else:
        return "Just Girls and Boys are available"

呼唤男生：

print(the_most_delicious_candy(my_dict, "boys"))

BOYS OUTPUT：

[{'Chocolate': {'Boys': 4, 'Girls': 3}}, {'Lollipops': {'Boys': 4, 'Girls': 2}}, {'gumdrop': {'Boys': 4, 'Girls': 3}}]

呼唤女孩：

print(the_most_delicious_candy(my_dict, "girls"))

GIRLS OUTPUT：

[{'gumdrop': {'Boys': 4, 'Girls': 3}}, {'Chocolate': {'Boys': 4, 'Girls': 3}}]

祝你好运......

Answer 4

这将有效：

var logoimg = document.getElementById("logo-cat"); 


    logoimg.src = "https://cdn.vox-cdn.com/uploads/chorus_asset/file/8597673/ph_1_color_black_moon_copy_1000x1000.png";

最大通话通过按键功能为您提供最多女孩的三连音。然后你就抓住那个三胞胎的糖果。

Answer 5

在执行max

时只需使用密钥

def get_candy_with_max_girls(favourite_candy_boys_girls):
    return max(favourite_candy_boys_girls, key = lambda k:k[2])[0]

返回第三列中具有最大值的子列表的第一项

5 个答案: