input id var1$  _20140303 _20140304;

cards;

1 价格 10 15

1 收入 11 17

2 价格 1 15

2 收入 2 15

;

run;

data b;set a;if id=1; 

data c;set a; if id =2;

proc transpose data=b out=d_1;

var var1  _20140303 _20140304;

copy id;

run;

proc transpose data=c out=d_2;

var var1  _20140303 _20140304;

copy id;

run;

data e1;set d_1; rename _name_=date col1=price col2=income ;label date="日期"; id=1;if col1='价格' then delete ;run;

data e2;set d_2; rename _name_=date col1=price col2=income ;label date="日期"; id=2;if col1='价格' then delete ;run;

data f;

set e1 e2;

label date= '日期' income='收入' price='价格';

run;

帮师弟处理数据时，遇到的一个问题。

首先看网络上的解决方案：

data score;  将 横向 数据变成 纵向 的结构。

input Student $9. +1 StudentID $ Section $ Test1 Test2 Final;

datalines;

Capalleti 0545 1 94 91 87

Dubose 1252 2 51 65 91

Engles 1167 1 95 97 97

Grant 1230 2 63 75 80

Krupski 2527 2 80 76 71

Lundsford 4860 1 92 40 86

Mcbane 0674 1 75 78 72

;

run;

proc transpose data = score out = score_transposed;

run;

proc print data = score_transposed;

run;

proc transpose data = score out = idnumber name = Test prefix = sn;

run;

proc print data = idnumber noobs;

run;

proc transpose data = score out = idlabel name = Test prefix = sn;

id studentid;

idlabel student;

run;

proc print data = idlabel label noobs;

run;

data fishdata;

infile datalines missover;

input Location & $10. Date date7.

Length1 Weight1 Length2 Weight2 Length3 Weight3

Length4 Weight4;

format date date7.;

datalines;

Cole Pond 2JUN95 31 .25 32 .3 32 .25 33 .3

Cole Pond 3JUL95 33 .32 34 .41 37 .48 32 .28

Cole Pond 4AUG95 29 .23 30 .25 34 .47 32 .3

Eagle Lake 2JUN95 32 .35 32 .25 33 .30

Eagle Lake 3JUL95 30 .20 36 .45

Eagle Lake 4AUG95 33 .30 33 .28 34 .42

;

proc transpose data = fishdata out = fishlength(rename = (col1 = Measurement));

var length1-length4;

by location date;

run;

proc print data = fishlength noobs;

run;

data stocks;

input Company $14. Date $ Time $ Price;

datalines;

Horizon Kites jun11 opening 29

Horizon Kites jun11 noon 27

Horizon Kites jun11 closing 27

Horizon Kites jun12 opening 27

Horizon Kites jun12 noon 28

Horizon Kites jun12 closing 30

SkyHi Kites jun11 opening 43

SkyHi Kites jun11 noon 43

SkyHi Kites jun11 closing 44

SkyHi Kites jun12 opening 44

SkyHi Kites jun12 noon 45

SkyHi Kites jun12 closing 45

;

proc transpose data = stocks out = close let;

by company;

id date;

run;

proc print data = close noobs;

run;

data weights;

input Program $ s1-s7;

datalines;

CONT 85 85 86 85 87 86 87

CONT 80 79 79 78 78 79 78

CONT 78 77 77 77 76 76 77

CONT 84 84 85 84 83 84 85

CONT 80 81 80 80 79 79 80

RI 79 79 79 80 80 78 80

RI 83 83 85 85 86 87 87

RI 81 83 82 82 83 83 82

RI 81 81 81 82 82 83 81

RI 80 81 82 82 82 84 86

WI 84 85 84 83 83 83 84

WI 74 75 75 76 75 76 76

WI 83 84 82 81 83 83 82

WI 86 87 87 87 87 87 86

WI 82 83 84 85 84 85 86

;

data split;

set weights;

array s{7} s1-s7;

Subject + 1;

do Time = 1 to 7;

Strength = s{time};

output;

end;

drop s1-s7;

run;

proc print data=split noobs;

run;

proc transpose data = split out = totsplit prefix = Str;

by program subject;

copy time strength;

var strength;

run;

proc print data = totsplit;

run;

师弟的例子：将 横向 数据变成 纵向 的结构。

PROC IMPORT OUT=weights

            DATAFILE= "D:\0218.txt"

            DBMS=TAB REPLACE;

     GETNAMES=YES;

     DATAROW=2;

RUN;

data split;

set weights;

array s{5} car1-car5;

Subject + 1;

do Time = 1 to 5;

car = s{time};

output;

end;

drop s1-s5;

run;

proc print data=split noobs;

run;

proc transpose data = split out = totsplit prefix = Str;

by   subject;

copy  time  v1 firm v3 v4 v5 v6 v7 v9 industry date v13 v14 v15 vcpe car;

var car;

run;

proc print data = totsplit;

run;

PROC EXPORT DATA=totsplit

            OUTFILE= "D:\02.csv"

            DBMS=CSV  REPLACE;

RUN;

PROC EXPORT DATA=split

            OUTFILE= "D:\01.csv"

            DBMS=CSV  REPLACE;

RUN;