from collections import *
def compress(s):
res = Counter(s)
for key,value in res.items():
print(key, value)
compress("hhgoogle")
输出:
h 2
g 2
o 2
l 1
e 1
如何根据计数按字母顺序排序，即:
所需输出:
g 2
h 2
o 2
e 1
l 1
首先按字母顺序排序，然后对出现的情况进行排序。 连接仅仅是以您所描述的格式输出它。
from collections import *
def compress(s):
res= Counter(s)
alphabetic_res = sorted(res.most_common(), key=lambda tup: tup[0])
final_res = sorted(alphabetic_res, key=lambda tup: tup[1], reverse=True)
print("\n".join("%s,%s" % tup for tup in final_res))
compress("hhgoogle")
OUT: g,2
h,2
o,2
e,1
l,1
您可以使用:
from collections import *
print(sorted(Counter('hhgoogle').most_common(), key=lambda x: (-x[1], x[0])))
# [('g', 2), ('h', 2), ('o', 2), ('e', 1), ('l', 1)]
演示
下面是您可以做的:
from collections import *
def compress(s):
res= Counter(s)
l1 = sorted(res.most_common(), key=lambda t: t[0])
l2 = sorted(l1, key=lambda t: t[1], reverse=True)
print('\n'.join([f"{t[0]} {t[1]}" for t in l2]))
compress("hhgoogle")
输出:
g 2
h 2
o 2
e 1
l 1
