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大家好，我有一个关于根据字典中的项目数打印最大值到最小值的问题。

我需要一个完整的密钥列表，上面印有密钥旁边显示的次数。

基于这两个列表：

Teams = ['Boston Americans','World Series Not Played in 1904','New York Giants',
         'Chicago White Sox','Chicago Cubs','Chicago Cubs','Pittsburgh Pirates',
         'Philadelphia Athletics']

 Year = [1903,1904,1905,1906,1907,1908,1909,1910]

如何根据每个团队拥有的项目数量打印以下内容：

D = {Teams,Year}

我忘了提到我正在从 txt 文件中读取这个。

更新 

def Team_data(team,year):
      D = defaultdict(list)
      for team,year in zip(team,year):
          D[team].append(year)
      pprint(D)
return D

然后它将返回：

D = {'Boston Americans':1903,'World Series Not Played in 1904':1904, 
 'New York Giants':1905,'Chicago White Sox': 1906,'Chicago Cubs': 1907,
 'Chicago Cubs':1908,'Pittsburgh Pirates':1909,'Philadelphia Athletics':1910}

Team_max = []

打印时，我在列表中得到以下内容

Chicago Cubs, 2
Boston Americans, 1
World Series Not Played in 1904, 1
New York Giants, 1
Chicago White Sox, 1
Pittsburgh Pirates, 1
Philadelphia Athletics, 1

我正试图从制作

团队 = 关键和年份 = 价值
Chicago Cubs:[1907,1908]

团队 = 键和显示次数 = 值
Chicago Cubs:[2]

我根据阅读的内容尝试使用以下内容：

def Team_data_max(D):
    key = D.keys()
    value = D.values()
    team_max = []
    team_max = sorted(D, key=lambda key: len(D[key]))
    print(team_max)
    #Print 1 Key based and 1 max number of items in dictionary
    #Chicago Cubs, 2

或者这个

    team_max = []
    team_max = max(((k, len(v)) for k, v in D.items()), key=lambda x: x[1])
    print(Series_max)
    #Print the entire key list least to greatest without a number which is what I need
    #based on the items list

然后这个

    s = []
    s = [k for k in D.keys() if len(D.get(k))==max([len(n) for n in D.values()])]
    print(s)
    #print max key without number
    #Chicago Cubs

虽然我在第二个函数中更接近了，但我只需要这个数字。关于如何解决这个问题的任何想法？任何想法或指导表示赞赏。

4

2 回答 2

1

你的意思是这样的吗？

from collections import defaultdict
Teams = ['Boston Americans','World Series Not Played in 1904','New York Giants',
         'Chicago White Sox','Chicago Cubs','Chicago Cubs','Pittsburgh Pirates',
         'Philadelphia Athletics']

Year = [1903,1904,1905,1906,1907,1908,1909,1910]
d=defaultdict(list)
for x,y in zip(Teams,Year):
    d[x].append(y)

for k,v in sorted(d.items(),key=lambda y:len(y[-1]),reverse=True):
    print "{0} {1}".format(k,",".join(map(str,v)))

输出：

Chicago Cubs 1907,1908
Chicago White Sox 1906
New York Giants 1905
World Series Not Played in 1904 1904
Philadelphia Athletics 1910
Pittsburgh Pirates 1909
Boston Americans 1903
于 2013-03-31T20:08:06.513 回答
1

不太确定你想在这里做什么。也许读一点听写？重复键-> 允许？这是我如何理解您的问题的解决方案：

Teams = ['Boston Americans','World Series Not Played in 1904','New York Giants',
         'Chicago White Sox','Chicago Cubs','Chicago Cubs','Pittsburgh Pirates',
         'Philadelphia Athletics']
Year = [1903,1904,1905,1906,1907,1908,1909,1910]

l = zip(Teams, Year)
d = {}
for e in l:
    if e[0] not in d:
        d[e[0]] = [e[1]]
    else:
        d[e[0]].append(e[1])

#Max Item
s = sorted(d.items(), key=lambda x: len(x[1]))[-1]
print("%s - %s" % (s[0], len(s[1])))

# Entire list
for k, v in sorted(d.items(), key=lambda x: -len(x[1])):
    print("%s - %s" % (k, len(v)))

如果您之后确实需要 dict，这是一个解决方案。请注意，排序将芝加哥按上面生成的列表长度放在最后一个位置。然后通过 [-1] 获取最后一项。

于 2013-03-31T20:31:27.080 回答