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python 统计数组中元素出现次数并进行排序的实例

程序员文章站 2024-02-09 11:31:58
如下所示: lis = [12,34,456,12,34,66,223,12,5,66,12,23,66,12,66,5,456,12,66,34,5,34]...

如下所示:

lis = [12,34,456,12,34,66,223,12,5,66,12,23,66,12,66,5,456,12,66,34,5,34]

def test1():
  #进行去重
  c = []
  for i in lis:
    if i not in c:
      c.append(i)
  #进行统计,生成二维列表
  b = []
  for i in c:
    num = 0
    for j in range(len(lis)):
      if lis[j] == i:
        num += 1
    a = []
    a.append(i)
    a.append(num)
    b.append(a)
  # 排序算法,按出现次数进行降序排列
  for i in range(len(b)):
    for j in range(i,len(b)):
      if b[i][1] < b[j][1]:
        temp = b[i]
        b[i] = b[j]
        b[j] = temp
  print(b)

def test2():
  # set进行去重,进行统计生成二维列表
  b = []
  for i in list(set(lis)):
    num = 0
    for j in range(len(lis)):
      if lis[j] == i:
        num += 1
    a = []
    a.append(i)
    a.append(num)
    b.append(a)
  # 排序算法,按出现次数进行降序排列
  for i in range(len(b)):
    for j in range(i,len(b)):
      if b[i][1] < b[j][1]:
        temp = b[i]
        b[i] = b[j]
        b[j] = temp
  print(b)

def test3():
  # 统计元素出现次数,元素为key,次数为value,生成字典
  a = {}
  for i in lis:
    if i in a:
      a[i] = a[i] + 1
    else:
      a[i] = 1
  # 使用sorted对字典进行排序
  b = sorted(a.items(),key=lambda item:item[1],reverse=True)
  print(b)

def test4():
  from collections import Counter
  import operator
  #进行统计
  a = dict(Counter(lis))
  #进行排序
  b= sorted(a.items(), key=operator.itemgetter(1),reverse=True)
  print(b)

if __name__ == '__main__':
  test1()
  test2()
  test3()
  test4()

输出结果如下:

[[12, 6], [66, 5], [34, 4], [5, 3], [456, 2], [223, 1], [23, 1]]
[[12, 6], [66, 5], [34, 4], [5, 3], [456, 2], [23, 1], [223, 1]]
[(12, 6), (66, 5), (34, 4), (5, 3), (456, 2), (23, 1), (223, 1)]
[(12, 6), (66, 5), (34, 4), (5, 3), (456, 2), (23, 1), (223, 1)]

这是面试过程中遇到的一个问题找到的解决方法,总结了一下,小编是初学者,还需不断努力学习。

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