我有一个这样的文件:
This is a file with many words.
Some of the words appear more than once.
Some of the words only appear one time.
我想生成一个由两列组成的列表。第一列显示出现的单词,第二列显示它们出现的频率,例如:
this@1
is@1
a@1
file@1
with@1
many@1
words3
some@2
of@2
the@2
only@1
appear@2
more@1
than@1
one@1
once@1
time@1
words
和word
可以算作两个单独的单词。到目前为止,我有这样的想法:
sed -i "s/ /\n/g" ./file1.txt # put all words on a new line
while read line
do
count="$(grep -c $line file1.txt)"
echo $line"@"$count >> file2.txt # add word and frequency to file
done < ./file1.txt
sort -u -d # remove duplicate lines
由于某些原因,这只在每个单词后显示"0“。
如何生成文件中出现的每个单词的列表以及频率信息?
发布于 2012-05-11 22:05:35
不是sed
和grep
,而是tr
、sort
、uniq
和awk
% (tr ' ' '\n' | sort | uniq -c | awk '{print $2"@"$1}') <<EOF
This is a file with many words.
Some of the words appear more than once.
Some of the words only appear one time.
EOF
a@1
appear@2
file@1
is@1
many@1
more@1
of@2
once.@1
one@1
only@1
Some@2
than@1
the@2
This@1
time.@1
with@1
words@2
words.@1
在大多数情况下,您还希望删除数字和标点符号,将所有内容转换为小写(否则"THE“、"The”和"the“将分别计算),并禁止输入长度为零的单词。对于ASCII文本,您可以使用以下修改后的命令执行所有这些操作:
sed -e 's/[^A-Za-z]/ /g' text.txt | tr 'A-Z' 'a-z' | tr ' ' '\n' | grep -v '^$'| sort | uniq -c | sort -rn
发布于 2014-10-03 06:37:17
uniq -c已经完成了您想要的操作,只需对输入进行排序:
echo 'a s d s d a s d s a a d d s a s d d s a' | tr ' ' '\n' | sort | uniq -c
输出:
6 a
7 d
7 s
发布于 2018-11-26 20:21:17
您可以使用tr来完成此任务,只需运行
tr ' ' '\12' <NAME_OF_FILE| sort | uniq -c | sort -nr > result.txt
城市名称文本文件的输出示例:
3026 Toronto
2006 Montréal
1117 Edmonton
1048 Calgary
905 Ottawa
724 Winnipeg
673 Vancouver
495 Brampton
489 Mississauga
482 London
467 Hamilton
https://stackoverflow.com/questions/10552803
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