問(wèn)題描述

例如此列表：

[[’software’, ’foundation’], [’of’, ’the’], [’the’, ’python’], [’software’, ’foundation’],[’of’, ’the’], [’software’, ’foundation’]]# 進(jìn)行頻率統(tǒng)計(jì)，例如輸出結(jié)果為：('[’software’,’foundation’]', 3), ('[’of’, ’the’]', 2), ('[’the’, ’python’]', 1)

問(wèn)題解答

# coding:utf8from collections import Countera = [[’software’, ’foundation’], [’of’, ’the’], [’the’, ’python’], [’software’, ’foundation’],[’of’, ’the’], [’software’, ’foundation’]]print Counter(str(i) for i in a) # 以字典形式返回統(tǒng)計(jì)結(jié)果print Counter(str(i) for i in a).items() # 以列表形式返回統(tǒng)計(jì)結(jié)果# -------------- map方法 --------print Counter(map(str, a)) # 以字典形式返回統(tǒng)計(jì)結(jié)果print Counter(map(str, a)).items() # 以列表形式返回統(tǒng)計(jì)結(jié)果回答2：

from collections import Counterdata = [[’software’, ’foundation’], [’of’, ’the’], [’the’, ’python’], [’software’, ’foundation’],[’of’, ’the’], [’software’, ’foundation’]]cnt = Counter(map(tuple, data))print(list(cnt.items()))回答3：

from itertools import groupbydata = ....print [(k, len(list(g)))for k, g in groupby(sorted(data))]