Question

loopCount = 0
candidateName = input("Please input the candidates name or type 'DONE' when finished:")
votes = []
while candidateName != "DONE":
     departmentName = input("Please input the department that submitted votes for this candidate:")
     numberOfVotes = input("Please input the number of votes for this candidate:")
     votes.append([])
     votes[loopCount].append(candidateName)
     votes[loopCount].append(departmentName)
     votes[loopCount].append(numberOfVotes)
     print(votes[loopCount])
     loopCount = loopCount + 1
     candidateName = input("Please input the candidates name or type 'DONE' when finished:")

print(votes)

sum = 0
for i in range(0,len(votes)):

这是我目前拥有的代码。我相信我需要做的就是用for循环来完成。我要做的是循环浏览列表，并累计每个候选人的投票数。当然，如果候选人具有匹配的名称，我需要正确分配。如果有人可以，请指出我如何以这种方式处理列表的正确方向。多维列表是什么让我失望。任何提示或建议都将不胜感激。谢谢。

Answer 1

所以，你要找的是一个关键的，价值结构你的数据，如python字典。由于这是二维（候选名称和部门名称），因此您需要一个嵌套字典。

Python不允许autovification像perl一样，所以你想使用defaultdict自动允许你像字典一样使用嵌套结构。

以下是重构版本：

import java.util.Scanner;   
public class StringShiftTwoLeftThenRight
{
    public static void main(String[] args)
    {
        String word, rightShift = "", leftShift = "";

        Scanner keyboard = new Scanner(System.in);
            System.out.print("\nEnter a word: ");
            word = keyboard.nextLine();
            rightShift = (word.substring((0),
            (word.length()-2)));

            leftShift = (word.substring((2),(word.length())));

            System.out.println("\nThe String shifted two to right looks like this: " + rightShift);
            System.out.println("\nThe String shifted two to left looks like this: " + leftShift);
            System.out.println("\nThe String as it is looks like: " + word);
    }
}

Answer 2

我的看法。有更好的方式，但它的工作原理

totalvotes = {}
for vote in votes:
    totalvotes[vote[0]] = totalvotes[vote[0]]+int(vote[2]) if vote[0] in totalvotes else int(vote[2])  
print(totalvotes)

或更紧凑

print([(gk,sum([int(v[2]) for v in gv] )) for gk, gv in itertools.groupby( sorted(votes,key=lambda x:x[0]),lambda x: x[0])])

怪异的单行：

function my_scripts_method() {
wp_enqueue_script(
    'custom-script',
    get_stylesheet_directory_uri() . '/masonry.php',
    array( 'jquery' )
);
}

add_action( 'wp_enqueue_scripts', 'my_scripts_method' );

Answer 3

我会使用python词典。根据需要修改代码。

votes = [['a','d',3],['b','d',3],['a','d',2]]

unique_names_list = {}
for ballot in votes:
    name = ballot[0]
    department = ballot[1]
    votes = ballot[2]

    name_depart = name + '-' + department

    if name_depart not in unique_names_list.keys():
        unique_names_list[name_depart] = votes
    else:
        unique_names_list[name_depart] = unique_names_list[name_depart] + votes

Answer 4

您只需要处理代码当前收集的结果。一种方法是这样的：

canidates = []
loopCount = 0
candidateName = input("Please input the candidates name or type 'DONE' when finished:")
votes = []
while candidateName != "DONE":
     departmentName = input("Please input the department that submitted votes for this candidate:")
     numberOfVotes = input("Please input the number of votes for this candidate:")
     votes.append([])
     votes[loopCount].append(candidateName)
     votes[loopCount].append(departmentName)
     votes[loopCount].append(numberOfVotes)
     print(votes[loopCount])
     loopCount = loopCount + 1
     candidateName = input("Please input the candidates name or type 'DONE' when finished:")

results = []

#Tally the votes
for v in votes:
    if v[0] in canidates:
        #need to add these votes to an existing canidate.
        for x in results:
            #goes through each record in results and adds the record to it.
            if x[0] == v[0]:
                x[1] = x[1] + ", " + v[1]
                x[2] = x[2] + int(v[2])
    else:
        #need to create a new record for the canidate.
        canidates.append(v[0])
        results.append([v[0],v[1],int(v[2])])


#Display the results
print ("Vote Results:")
for x in results:
    print ("name: " + x[0])
    print ("Departments: " + x[1])
    print ("Votes Total: " + str(x[2]))

因此，当同一个canidate名称有多个条目时，您要做的就是完成并合并条目。

Python二维列表处理

4 个答案: