统计字符串出现最多的字母及其出现次数
另外如果次数相同按字母顺序排序。
方法1
可以使用自定义键对c.most_common()进行排序，该键首先考虑频率的降序，然后考虑字母的降序（请注意lambda x: (-x[1], x[0]) ）：
from collections import Counter
def ordered_letters(s, n=3):
c = Counter(s.replace(' ', ''))
top_n = sorted(c.most_common(), key=lambda x: (-x[1], x[0]))[:n]
for i, t in enumerate(top_n):
c, f = t
if i == 0: print('1st most frequent', c + '.', 'Appearances:', f)
elif i == 1: print('2nd most frequent', c + '.', 'Appearances:', f)
elif i == 2: print('3rd most frequent', c + '.', 'Appearances:', f)
else: print(str(i + 1) + 'th most frequent', c + '.', 'Appearances', f)
sent = "china construction bank"
ordered_letters(sent, 5)
# 1st most frequent n. Appearances: 4
# 2nd most frequent c. Appearances: 3
# 3rd most frequent a. Appearances: 2
# 4th most frequent i. Appearances 2
# 5th most frequent o. Appearances 2
方法2
常规方式对Counter的元组进行排序， 但第一个参数-count本身被取反。 这将产生一个反向列表，但第二个元组元素将按字母顺序排序。 然后取最后n个项目。
from collections import Counter
ordinal = lambda n: "%d%s" % (n,"tsnrhtdd"[(n/10%10!=1)*(n%10<4)*n%10::4])
def ordered_letters(s, n=3):
ctr = Counter(c for c in s if c.isalpha())
ctr = sorted(ctr.items(), key=lambda x: (-x[1], x[0]))[:n]
for index,value in enumerate(ctr):
print "{:s} most frequent: '{:}'. Appearances: {:}".format(ordinal(index+1),value[0],value[1])
s = "achina aconstruction banck"
ordered_letters(s, n=3)
结果：
1st most frequent: 'a'. Appearances: 4
2nd most frequent: 'c'. Appearances: 4
3rd most frequent: 'n'. Appearances: 4
要想成为python这领域的高手，必须了解和弄懂python常见问题。要我们来进行python高手修炼吧。或者你有更好的问题可以留言。
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