删除csv文件中第一列具有6位数字的行

Question

我有一个这样的csv文件

"5478",a,56.40,-0.40 ,55.50,57.50,55.30,56.74,"862,971","48,962,460","695",56.40,56.60,"127,474,332",56.40,60.30,52.50

"5480",b,21.90,-0.25 ,21.80,22.00,21.80,21.87,"1,598,041","34,950,597","590",21.90,21.95,"199,097,830",21.90,23.40,20.40

"70462P",c,0.01,-0.01 ,0.01,0.01,0.01,0.01,"99,000","990","1",0.01,0.06,"5,000,000",0.01,0.31,0.01

"70465P",d, ---,--- ,---,---,---,0.02,"0","0","0",0.01,0.03,"20,000,000",0.02,0.32,0.01

"8935",bt,5.02,-0.02 ,4.95,5.19,4.92,5.05,"949,102","4,791,070","290",5.02,5.07,"201,902,107",5.02,5.37,4.67

1333,tnd,21.40,-0.60 ,22.00,22.20,21.20,21.52,"1,519,292","32,692,804","631",21.40,21.50,"102,525,625",21.40,22.85,19.95

我想检查第一列（如果超过4位数字），然后删除该行，例如，第二行和第三行将被删除。我该怎么做呢？多谢

ps2这是从股票中心下载的股票信息，但是我发现格式最近发生了变化，之前的格式是最后一行，第一列没有引号“”，是否可以过滤这两种格式？还是我应该处理这两种情况？

Answer 1

这是一个sed解决方案：sed -e '/^"[0-9]\{5\}/d' in-file > out-file您还可以使用-i选项进行就地替换：sed -i -e '/^"[0-9]\{5\}/d' file

Answer 2

不确定您要使用哪种语言，因为您标记了awk并在标签中添加了sed，但是您可以简单地使用grep：egrep '^\"[0-9]{1,4}\"' file.txt

Answer 3

awkawk -F, '$1~ /^\"[0-9][0-9]?[0-9]?[0-9]?\"$/' fileGNU sedsed '/^\"[0-9]\{1,4\}\"/!d' file