Question

我想弄清楚如何计算两个合并的名单列表的得分。我需要为每个字符（包括名字和姓氏之间的空格）加一个点，并为名称中的每个元音添加一个点。

<%= form_tag(products_path, :method => "get", id: "search-form") do %>
    <%= text_field_tag :search, params[:search], placeholder: "Search products" %>
    <%= submit_tag "Search", :name => nil %>
<% end %>

我正在努力弄清楚要做什么。任何帮助将不胜感激。

Answer 1

首先是zip名字和姓氏，然后以str为' '作为分隔符。然后用collections.Counter计算所谓的元音字符出现次数，sum它们，并添加全名len。那将是dict对象，然后你可以随心所欲地做任何事情。

from collections import Counter

a = ["John", "Kate", "Oli"]
b = ["Green", "Fletcher", "Nelson"]    
vowel = ["a", "e", "i", "o", "u"]

output = {}
for item in [' '.join(i) for i in zip(a,b)]:
    output[item] = sum(Counter(item)[x] for x in vowel) + len(item)
output

输出：

{'John Green': 13, 'Kate Fletcher': 17, 'Oli Nelson': 13}

<强>更新

如果您需要名字和姓氏的所有可能变体，则可以使用itertools.product

执行此操作

from itertools import product
from collections import Counter

a = ["John", "Kate", "Oli"]
b = ["Green", "Fletcher", "Nelson"]    
vowel = ["a", "e", "i", "o", "u"]

output = {}
for item in [' '.join(i) for i in product(a,b)]:
    output[item] = sum(Counter(item)[x] for x in vowel) + len(item)
output

输出：

{'John Fletcher': 16,
 'John Green': 13,
 'John Nelson': 14,
 'Kate Fletcher': 17,
 'Kate Green': 14,
 'Kate Nelson': 15,
 'Oli Fletcher': 15,
 'Oli Green': 12,
 'Oli Nelson': 13}

Answer 2

for first, second in gen:
    name = " ".join(first, second)
    score = 0
    for letter in name:
        if letter in vowel:
            score += 1
    score += len(name)

将它们作为字符串更容易，然后可以循环遍历该字符串。每个字符一个点很容易，这只是字符串的长度。如果元音是一个额外的字母，只需循环字母，如果元音添加一个点。瞧！

计算合并字符串的得分

2 个答案: