Question

我刚开始在学校使用python，我有一个问题，我一直试图找出一段时间

问题是按频率排序列表，列表也包含字符串对于前给定的函数调用

SortByFrequency(['pie', 6, 'pie', 9, 6, 7, 9, 9]

它应该返回

[9, 9, 9, 'pie', 'pie', 6, 6, 7]

我如何使用python感谢找到解决方案我已经尝试过的代码是尝试使用字典并以某种方式打印元素

my_Dict ={}
for i in mylist:
       if i not in my_dict:
and count the occurrences

Answer 1

如果这不是某种不允许使用python模块的学校作业，请不要重新发明轮子，可以使用集合模块完成以下操作

import collections
def SortByFrequency(lst):
  return list(collections.Counter(lst).elements())

SortByFrequency(['pie', 6, 'pie', 9, 6, 7, 9, 9])
# this return [9, 9, 9, 'pie', 'pie', 6, 6, 7]

我自己尝试用字典来解决这个问题

def SortByFrequency(mylist):
    my_dict = {}
    for i in mylist:
        my_dict[i] = my_dict.get(i,0) + 1
    return sorted(sorted(mylist,key=str), key=my_dict.get, reverse=True)

SortByFrequency(['pie', 6, 'pie', 9, 6, 7, 9, 9])
# but this does not guarantee the order when we have multiple values with same frequency
# this will return [9, 9, 9, 6, 6, 'pie', 'pie', 7]

Answer 2

你正在建立字典的路上。像这样完成它：

$doc = [xml] (Get-Content -raw gamelist.xml)

foreach ($gameEl in $doc.DocumentElement.game) { 
  # Use -replace to extract the filename without extension from the 
  # path contained in the <path> element.
  $gameName = $gameEl.path -replace '^.*/(.*)\..*$', '$1'
  # Append elements 'video' and 'marquee', but only if they don't already
  # exist.
  if ($null -eq $gameEl.video) {
    $gameEl.AppendChild($doc.CreateElement('video')).InnerText = "~/.emulationstation/roms/megadrive/Videos/${gameName}.mp4"
  }
  if ($null -eq $gameEl.marquee) {
    $gameEl.AppendChild($doc.CreateElement('marquee')).InnerText = "~/.emulationstation/roms/megadrive/Marquees/${gameName}.png"
  }
}

$writer = [System.IO.StreamWriter] "$PWD/newFile.xml"
$doc.Save($writer)
$writer.Close()

然后你只需要使用字典作为键对其进行排序：

    if i not in my_dict:
        my_dict[i] = 0 # define this entry
    my_dict[i] += 1 # increment it (number of occurrences of i)

减号是一种按降序排序的快捷方式。请注意，使用初始小写字母编写函数更为常见，因为初始资本通常保留给类名。

Answer 3

您必须使用计数器

创建辅助字典

list_ = ['pie', 6, 'pie', 9, 6, 7, 9, 9]
dict_ = {}
for i in list_:
  dict_[i] = dict_.get(i, 0) - 1
# Your dict_ now is following:
# {6: -2, 7: -1, 9: -3, 'pie': -2}
sorted(list_, key=dict_.get)
#=> [9, 9, 9, 'pie', 6, 'pie', 6, 7]

Python排序频率

3 个答案: