Question

我正在尝试编写一个函数，该函数返回一个列表，其中包含名称在文件中出现的次数。我有两个帮助程序功能，它们遍历文件并获得两个列表，一个带有所有名称，然后另一个带有所有唯一名称。我想使用唯一名称并将它们与所有名称进行比较，并计算唯一名称出现在所有名称列表中的次数。

    #call getAllPeople helper function to get list
    allName=getAllPeople(fileName)
    #call getUniquePeople from helper function to get comparison
    uniNam=getUniquePeople(fileName)
    #give empty list
    namNum=[]
    #initiate counter
    count=0
    #iterate through list
    for name in allName:
        #if name is in getUniquePeople(fileName)
        if name in uniNam:
            #add to count
            count = count+1

            return count

我正在寻找：

['bob:4', 'jane:4', 'pam:2', 'batman:1']

在和它一起玩时：

#give empty list 
namCount=[] 
#initiate counter 
count=0 
#iterate through list 
for name in range(len(allName)): 
    #if name is in getUniquePeople(fileName) 
    if name in uniNam: 
    #add to count 
    count = count+1 
    #append namCount with 
    namCount.append(name +':'+str(count)) 
    return namCount

我一无所获

Answer 1

正如@Hoog所说，字典更适合解决这个问题。

# make a new empty dictionary
results = {}

for name in allName:
    if name in uniNam:
        # if name is already in the results dictionary, add 1 to its count
        if name in results:
            results[name] = results[name] + 1
        # otherwise create a new name in the dictionary with a count of 1
        else:
            results[name] = 1

{'bob': 4, 'jane': 4, 'pam': 2}

编辑-如果您绝对必须只使用列表：

# make new lists for holding names and counts
names = []
name_counts = []

for name in allName:
    if name in uniNam:
        # if name is already in our list, add 1 to the the corresponding entry in the name_counts list
        if name in names:
            position = names.index(name)
            name_counts[position] = name_counts[position] + 1
        # otherwise add this name to our lists with a count of 1
        else:
            names.append(name)
            name_counts.append(1)

# now combine our two lists into the final result
final_list = []
for position in range(len(names)):
    final_list.append(names[position] + ':' + str(name_counts[position]))

Answer 2

您可以使用Counter实现此目的。

from collections import Counter
names = ['John G.', 'Mark T.', 'Brian M.', 'Mark T.', 'Kit H.'] # All names
c = Counter(names)
print(c)

Output:
Counter({'Mark T.': 2, 'John G.': 1, 'Brian M.': 1, 'Kit H.': 1})

计算字符串出现在列表中的次数

2 个答案: