在python中不使用字典计算词频

Question

我有一个按字母顺序排序的文件输入

Asburn 9
Ashley 10
Ashley 11
Bojourn 12

我希望输出像

Asburn 9 
Ashley 21 
Bojourn 12 

我用过的代码:

counts = defaultdict(int)
for line in sys.stdin:
word, count  = line.split()
counts[word] += int(count)
for wrd, count in counts.items():
print("{}t{}".format(wrd,count))

但是我不会用字典。我应该如何实现它

Answer 1

通过Python标准库的帮助:

from sys import stdin
from itertools import groupby
from operator import itemgetter
for word, group in groupby(map(str.split, stdin), itemgetter(0)):
print(word, sum(int(count) for _, count in group))

上网试试!

Answer 2

试试这个:

prev_word = None
prev_count = 0
for line in sys.stdin:
word, count  = line.split()
count = int(count)
if word != prev_word:
if prev_word is not None:
print(prev_word, str(prev_count))
prev_count = count
prev_word = word
else:
prev_count += count
if prev_word != None:
print(prev_word, prev_count)

快速测试:下面的代码返回预期的输出

prev_word = None
prev_count = 0
for line in ['ashutosh 9', 'ashutosh 10', 'check 11', 'check 11', 'zebra 11']:
word, count  = line.split()
count = int(count)
if word != prev_word:
if prev_word is not None:
print(prev_word, prev_count)
prev_count = count
prev_word = word
else:
prev_count += count
if prev_word != None:
print(prev_word, prev_count)

Answer 3

使用正则表达式列出类似键值的术语，按键分组，将每组的键值与zip解耦

import re
import itertools as it
text = """Asburn 9
Ashley 10
Ashley 11
Bojourn 12"""
pairs = re.findall(r'^(.+?)s(d+)n*', text, re.M)
groups = [zip(*grp) for _, grp in it.groupby(pairs, lambda p: p[0])]
for keys, values in groups:
print(keys[0], sum(map(int, values)))