版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/u010870545/article/details/52652706
def most_common(seq):
d = {}
for i in seq:
d[i] = d.get(i, 0) + 1
ret = []
for j in sorted(d.items(), reverse=True, key=lambda x:x[1]):
if len(ret) == 0:
ret.append(j[0])
n = j[1]
else:
if j[1] == n:
ret.append(j[0])
else:
break
return ret
print '频次最高', most_common([1,2,3,4,5,1,1,2,2])