请教:如何把下面数据中第四位的点去掉(为列数据),并且在复制到stata中时前面的0仍然保留。
001.19 001.21 001.22 001.39 001.41 001.49 001.51 001.52 001.9 025.3 041.1 041.2 042.1 042.2 043.0 045.1 045.3 045.91 045.92 045.93 045.99 054.84 054.85 054.87 054.88 054.89 057.71
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楼主: honestwh
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2466
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请教数据处理 |
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已卖:1929份资源 教授 82%
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回帖推荐另一种处理方法:clearinput str6 x001.19001.21001.22001.39001.41001.49001.51001.52001.9025.3041.1041.2042.1042.2043.0045.1045.3045.91045.92045.93045.99054.84054.85054.87054.88054.89057.71endsplit x, parse(.)gen y = x1+x2*=========结果===============. list +---------------------------+ | x x1 ...
/*录入数据*/input str6 x001.19001.21001.22001.39001.41001.49001.51001.52001.9025.3041.1041.2042.1042.2043.0045.1045.3045.91045.92045.93045.99054.84054.85054.87054.88054.89057.71end/*替换小数点*/local h=length(x)forvalues i=1/`h' {replace x=substr(x,1,`i'-1)+substr(x,`i'+1,`h'-`i') if substr(x,`i',1)=="." }/*结束*/
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