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私は連想リストで単語の繰り返しを見つけるためのスクリプトを持っています:

newm = [u'life', u'selection', u'squire', u'naturalist', u'patriarch', u'home', u'man', u'public', u'nbsp', u'born', u'naturalist', u'theory', u'selectionbecame', u'foundation', u'country', u'gentleman', u'suggesting', u'class', u'time', u'death', u'evolutionary', u'imagery', u'ofscience', u'literature']
print newm

#count for list
counts = defaultdict(int)
print "uyti"
for x in newm:
    counts[x]+=1

print counts

このプログラムは "uyti" も出力しません。エラーは何ですか？

4

3 に答える 3

0

Python 2.5以降を想定

from collections import defaultdict

newm = [u'life', u'selection', u'squire', u'naturalist', u'patriarch', u'home', u'man', u'public', u'nbsp', u'born', u'naturalist', u'theory', u'selectionbecame', u'foundation', u'country', u'gentleman', u'suggesting', u'class', u'time', u'death', u'evolutionary', u'imagery', u'ofscience', u'literature']
print newm

#count for list
counts = defaultdict(int)
print "uyti"
for x in newm:
   counts[x]+=1

print counts

defaultdictを作成し、カウントとキーを入力して印刷を行います（これはデバッグ/サニティチェックのものだと思います...）

于 2013-08-20T16:12:59.097 に答える
0 
from collections import defaultdict
newm = [u'life', u'selection', u'squire', u'naturalist', u'patriarch', u'home', u'man', u'public', u'nbsp', u'born', u'naturalist', u'theory', u'selectionbecame', u'foundation', u'country', u'gentleman', u'suggesting', u'class', u'time', u'death', u'evolutionary', u'imagery', u'ofscience', u'literature',u'home']

#count for list

counts = defaultdict(int)

for x in newm:
   counts[x]+=1

print counts
于 2013-08-20T16:12:11.020 に答える