Python二維列表處理

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)):

這是我目前擁有的代碼。 我相信我需要做的事可以通過for循環來完成。 我想做的是循環瀏覽列表，並將每個候選人的投票總數加起來。 當然，如果候選人的姓名匹配，我將需要對其進行正確分配。 如果有人可以，請為我如何以這種方式處理列表指明正確的方向。 多維列表讓我很吃驚。 任何提示或建議都將不勝感激。 謝謝。

Answer 1

因此，您正在尋找的是鍵值結構的數據，例如python字典。 由於這是二維的（候選人名稱和部門名稱），因此需要一個嵌套詞典。

Python不容許autovification如Perl，所以要使用defaultdict自動允許您使用嵌套結構就像一本字典。

這是重構版本：

from collections import defaultdict
candidates_and_votes = defaultdict(dict)
while True:
     candidateName = input("Please input the candidates name or type 'DONE' when finished:")
     if candidateName == 'DONE':
         break
     departmentName = input("Please input the department that submitted votes for this candidate:")
     numberOfVotes = input("Please input the number of votes for this candidate:")
     candidates_and_votes[candidateName][departmentName] = numberOfVotes

for candidate, department in candidates_and_votes.items():
    total = 0
    print('Candidate Name {}'.format(candidate))
    for dep_name, number in department.items():
        print('  Dep {}. total {}'.format(dep_name, number))
        total += int(number)
    print('     Total Votes = {}'.format(total))

Answer 2

我拿 有更奇妙的方法，但是可以

totalvotes = {}
for i in range(0,len(votes)):
     if not votes[i][0] in totalvotes:
          totalvotes[votes[i][0]] = 0
     totalvotes[votes[i][0]] = totalvotes[votes[i][0]]+int(votes[i][2])
print(totalvotes)

或更緊湊

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)

怪異的一線：

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])])

Answer 3

我會用python字典。 根據需要修改您的代碼。

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

您只需要處理代碼當前正在收集的結果。 一種方法是這樣的：

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，當有多個候選名稱相同的條目時，您要做的就是遍歷並合並這些條目。