python求众数问题实例
本文实例讲述了python求众数问题的方法，是一个比较典型的应用。分享给大家供大家参考。具体如下：
问题描述:
多重集中重数最大的元素称为众数...就是一个可以有重复元素的集合,在这个集合中重复的次数最多的那个数就叫它的众数...
如S = [1,2,2,2,3,5] 重数是2,其重数为3
实例代码如下:
list_num = []
list_num_count = 0
dict_num ={}
#从文件读入,文件第一行为集合中元素的个数,以后每一行为一个元素
list_num_count = int(open('input.txt','r').readline())
for line_num, line in enumerate(open("input.txt",'r')):
if line_num > 0:
list_num += line.split()
#将读到的元素加入的字典中
for item in list_num:
if dict_num.has_key(item):
dict_num[item] += 1
else:
dict_num.setdefault(item,1)
pass
#找到出现次数最多的那个数,找到重数
dict_sort_by_top = {}
top_value = 0
for valus in dict_num.itervalues():
if valus> top_value:
top_value = valus
pass
#根据重数找到众数...这是因为考虑到可能有多个元素有相同多的重数
the_pop_num = 0
the_pop_num_count = 0
for keys,values in dict_num.iteritems():
if values == top_value:
print 'the pop num is %s,and the appear num is %s' % (keys,values)
the_pop_num = keys
the_pop_num_count = values
#输出到文件,第一行为从数,第二行为重数
write_line = '%s\n%s' %(the_pop_num, the_pop_num_count)
open("output.txt",'w').write(write_line)
这里假设有同级目录文件input.txt内容如下：
8 11 37 2 37 2 45 99 37
第一行的8代表元素个数，其后每一行有一个元素。
测试环境为Python2.7.6，
Python程序针对input.txt文件操作的运行结果如下：
the pop num is 37,and the appear num is 3
同时生成output.txt文件记录了众数37及其重复次数3。
希望本文所述对大家的Python程序设计有所帮助。
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