t********m
发帖数: 939 | 1 原data set如下：
ID Session var1 var2 var3 var4
1 1 10 . .
1 1 . 20 .
1 1 . . 30
1 1 . . . black
1 2 11 . .
1 2 . 22 .
1 2 . . . white
想要得到的data set如下：
ID Session var1 var2 var3 var4
1 1 10 20 30 black
1 2 11 22 . white
所用的程序如下：
proc sql;
create table short as
select ID, Session,
max(var1) as var1,
max(var2) as var2,
max(var3) ... 阅读全帖 |
|
H*******r
发帖数: 98 | 2 one is...
data select;
set original;
where var1 in (value1 value2 value3 value4 value5 value6 value7 value8)
or
var1 between minvalue and maxvalue or
var2 in (number1 number2 number3) or
var3 in (number1 number2 number3) or
var4 in (number1 number2 number3) or
var5 in (number1 number2 number3) or
var6 in (number1 number2 number3) or
var7 in (number1 number2 number3) or
var8 in (number1 number2 number3) or
... 阅读全帖 |
|
k*******a
发帖数: 772 | 3 you can create a score for each obs, then find the maximum score, for
example
data test;
input id var1 $ var2 $ var3$;
datalines;
1 A B B
1 B C B
1 A B C
2 A A B
2 A A B
2 C C B
3 C A A
3 B A A
3 A A A
;
proc sql;
select *, 100*((var3='C')*3 + (var3='B')*2 + (var3 = 'A'))
+ 10*((var2='C')*3 + (var2='B')*2 + (var2 = 'A'))
+ ((var1='C')*3 + (var1='B')*2 + (var... 阅读全帖 |
|
l****u
发帖数: 529 | 4 proc sql;
create table two as
select *, case when min(var3)=max(var3) then ' ' else max(var3) end as id3,
case when min(var2)=max(var2) then ' ' else max(var2) end as id2,
case when min(var1)=max(var1) then '' else max(var1) end as id1
from test
group by id
having var3=id3 or(id3=' 'and var2=id2) or (id3=''and id2=''and var1=id1);
quit; |
|
d*********d
发帖数: 239 | 5 不知道有没有理解对你的意思。
%MACRO Test();
PROC SQL;
SELECT * FROM
A
HAVING Var1 = MIN(Var1) OR Var1 = MAX(Var1);
QUIT;
%MEND Test;
%Test;
）。 |
|
w******d
发帖数: 120 | 6 比如var1
Range Value
1-5 --- 3
6-7 --- 8
8-11 --- 20
...
Var2
Chicago --- 10
New york --- 15
如果数据很大， 变量多， 用if- else 很慢
if var1 >= 1 and var1 <5:
value = 3
elif var >=6 and var1 < 7:
value = 8
....
if var2 = 'chicago':
value = 10
Python 有没有好的函数做这个？
最好能handle所有情况， 而不是每个var都不一样 |
|
t**********r
发帖数: 182 | 7 Many thanks for your hint!! I made it. Here is the code.
579 proc sql;
580 create table data3 as
581 select data1.*, data2.rating, day1-day2 as diff
582 from data1, data2
583 where data1.var1=data2.var1 and data1.var2=data2.var2
584 and date1-date2>0
585 group by data1.var1, data1.var2, data1.date1
586 having diff=min(diff);
NOTE: The query requires remerging summary statistics back with the original
data.
NOTE: Table WORK.data3 created, with 48144 rows and 9 |
|
w********p
发帖数: 948 | 8 问题继续， 包子继续
class Base
{
public:
Base(){ cout<<"Constructor: Base"<
~Base(){ cout<<"Destructor : Base"<
};
class Derived: public Base
{
public:
Derived(){ cout<<"Constructor: Derived"<
~Derived(){ cout<<"Destructor : Derived"<
};
int main()
{
Derived Var1;
Base Var2(Var1);
Derived Var3(Var1);
return 0;
}
output：
Constructor: Base
Constructor: Derived
Destructor : Derived
Destructor : Base
Destructor : Base
Destructor : Derived
Destructor : Base
请解释一下 |
|
q******r
发帖数: 6542 | 9 ☆─────────────────────────────────────☆
handsomex (潇潇细雨) 于 (Thu Mar 24 23:05:47 2011, 美东) 提到:
有什么好的度量吗？
☆─────────────────────────────────────☆
benbenma (猪猪) 于 (Thu Mar 24 23:34:13 2011, 美东) 提到:
regression？就是那个啥r2的东东？
☆─────────────────────────────────────☆
handsomex (潇潇细雨) 于 (Thu Mar 24 23:40:23 2011, 美东) 提到:
这是什么啊？
☆─────────────────────────────────────☆
benbenma (猪猪) 于 (Thu Mar 24 23:42:47 2011, 美东) 提到:
统计分析啊？
☆─────────────────────────────────────☆
handsomex (潇潇细雨) 于 ... 阅读全帖 |
|
b********1
发帖数: 291 | 10 你们都是聪明人，谁能帮忙写个例子。 就算是1对多的join.
a的变量假设是var1, var2,var3
b的变量假设是var4,var5,var6
假设var1,var4是primary key, foreign key的关系。
不管是用scala, python 还是rdd spark.
我打算先看懂哪个就学哪个.
本人编程零基础, 只会写query.
说穿了，我就是想要 create table c as
select a.*, b.var5, b.var6
from a
join b
on a.var1=b.var4
然后把c下载到excel里面看看.
网上看了半天教程， 都是天马行空的东西。
我就奇了怪了， 这么简单的事情，hadoop上怎么就这么难实现？ |
|
w****c
发帖数: 10 | 11 one question about regular expression:
assuming we have statement as following:
%var1%/dir:%var2%/dir:%var3%/dir
in vi, it is expected that using s/%\(.*\)%/$\1/g the above
could be converted to "$var1/dir:$var2/dir:$var3/dir".but
I found that it wound up with "$var1%/dir:%var2%/dir:%var3/dir",
that is, vi searchs for largest matchup. How could I get the
fewest match up?
it is very interesting question. hope unix guru could help me.
Acton |
|
s*****n
发帖数: 2174 | 12 1. names(data)[1] <- "newname" 就可以, 如果你不喜欢用数字index, 也可以这样
names(data)[names(data)=="var1"] <- "newname" 或者
names(data) <- gsub("var1", "newname", names(data)) 都可以
2. 你说那个有个条件, 就是BY variable必须是相同的. 考虑如果data1, data2,
data3之间做一个merge. data1和data2之间用var1和var2来做index match, 而data1和
data3之间用var3来做index match. 反正就是这种比较复杂的merge, 每个data之间的
BY variable都不确定. 很难定义一个函数来handle多个data, 除非这个函数本身提供
很多很多参数.
3. 除了SAS, 还有别的语言有你说的这种"最近的data"的概念吗?
是最近一个赋值(写)的, 还是最后一个取值(读)的? 比如
data3 <- merge(data1, data2)
print(data2 |
|
o******6
发帖数: 538 | 13 ☆─────────────────────────────────────☆
lihao (思君) 于 (Wed Feb 13 10:12:51 2008) 提到:
data aa;
input id var1 var2;
cards;
001 1 1
002 1 2
003 3 4
;
I want to calculate the var1 sum of all 3 cases.
please help! thanks!
☆─────────────────────────────────────☆
lihao (思君) 于 (Wed Feb 13 10:22:40 2008) 提到:
☆─────────────────────────────────────☆
wuhuya (wuhuya) 于 (Wed Feb 13 11:27:47 2008) 提到:
to many ways to do this. just listed two of them
1. proc means data=aa;
var var1;
o |
|
t**********r
发帖数: 182 | 14 Has figured it out. Thanks.
===============
579 proc sql;
580 create table data3 as
581 select data1.*, data2.rating, day1-day2 as diff
582 from data1, data2
583 where data1.var1=data2.var1 and data1.var2=data2.var2
584 and date1-date2>0
585 group by data1.var1, data1.var2, data1.date1
586 having diff=min(diff);
NOTE: The query requires remerging summary statistics back with the original
data.
NOTE: Table WORK.data3 created, with 48144 rows and 9 columns.
587 quit;
NOTE: PROCEDURE SQL used (Tota |
|
y******0
发帖数: 401 | 15 proc sql;
create table data3 as
select data1.var1,data1.var2, data2.rating, min(date1-date2) as diff
from data1, data2
where data1.var1=data2.var1
and data1.var2=data2.var2
and date1>date2
group by 1,2,3;
quit; |
|
x*******u
发帖数: 500 | 16 比如说我有一个data;
ID var1 var2
1 0.3 0.4
2 0.5 0.6
3 0.1 5
想实现if id=1 var1=(id=2时 var2的值),
也就是这样
ID var1 var2
1 0.6 0.4
2 0.5 0.6
3 0.1 5
请指教. 谢谢 |
|
D******n
发帖数: 2836 | 17 proc sort data=a1;by var1 ;run;
data a2 (drop = i);set a1;by var1;retain i;
if (first.var1 eq 1) then i=0;i=i+1;
if (i<=10) then output;run; |
|
y*********4
发帖数: 76 | 18 I could use that successfully without macro as:
if (first.var1) then i=0;
but when I tried to write that in a macro
%if (first.var1) %then i=0;
it runs with error information: A character operand was found in the %EVAL
function or %IF condition where a numeric
operand is required. The condition was: (FIRST.RATERID)
btw, both var1 and i are not macro variables.
Is there anyone who know how to solve it?
Thanks a lot! |
|
d*******1
发帖数: 854 | 19 例子
ID VAR1 VAR2 VAR3
1 1 2 3
2 4 5 6
转化成:
ID VARNAME VALUE
1 VAR1 1
1 VAR2 2
1 VAR3 3
2 VAR1 4
2 VAR2 5
2 VAR3 6
column |
|
a***r
发帖数: 420 | 20 抛砖引玉
#original matrix: A
#new matrix: X
#number of obs:n
#number of var:nvar
> A
var1 var2 var3 var4 var5
1 1 2 3 4 5
2 6 7 8 9 10
3 11 12 13 14 15
4 16 17 18 19 20
>n=4
>nvar=5
>value <-as.vector(t(A))
>varname <-rep(colnames(A),n)
>ID <-sort(rep((1:n),nvar))
>X <-data.frame(cbind(ID,varname,value))
ID var value
1 1 var1 1
2 1 var2 2
3 1 var3 3
4 1 var4 4
5 1 var5 5
6 2 var1 6
7 2 var2 7
8 2 var3 |
|
d*******1
发帖数: 854 | 21 比如：
ID1 ID2 var1 var2 var3
A 1 10 44 1
A 1 10 44 2
A 2 11 3 3
A 2 11 3 4
B 3 12 4 5
B 3 12 4 6
B 4 20 10 7
作为一个DATAFRAME, 转化成如下：
ID1 ID2 var1 var2 var3
A 1 10 44 1
A 2 11 3 3
B 3 12 4 5
B 4 20 10 7
换句话说， var1 and var2 的值对每一个ID1xID2都是unique的， var取第一个值就可
以了。
用了aggregate好像不赶趟：
test<-aggregate(data[,3:4], list(data[,1:2]), mean)
Error in sort.list(y) :
'x' must be atomic for 'sort.li |
|
s******r
发帖数: 1524 | 22 proc sort data=test;by var1 descending var3 ;run;
data t1;
set test;
by var1;
if first.var1 then do;
if var3>0 then varx=var3;else varx=.;
end;
var4=max(var3,varx);
retain varx;
drop varx;
run;
|
|
G*****u
发帖数: 1222 | 23 我现在一般用sql 好久没用sas sql了
sql语句应该是 （假设大数据是large,小数据是small,共同的variable是A
select var1, var2, var3,
case when A in (select A from small) then 'Patient' else 'Non-Patient' end
as indicator
from large
group by var1, var2, var3
having case when A in (select A from small) then 'Patient' else 'Non-Patient
' = 'Patient'
这个是病人的数据
把最后一句换成
having case when A in (select A from small) then 'Patient' else 'Non-Patient
' = 'Non-Patient'
就是非病人的数据
或者简单一点 不多加变量的话
你可以用个sub query来得到病人的数据
select var1, var2, var3
from la |
|
S*x
发帖数: 705 | 24 select *
from data1 as a
left join data2 as b on (a.var1=b.var1)
where b.var1 is null
VAR2, |
|
o****o
发帖数: 8077 | 25 你这个用BY statement在SAS里面也不难，分条件运行不同回归也很容易。比如
data original;
array _x{*} var1-var10;
do i=1 to 10000;
do _j=1 to dim(_x); _x[_j]=rannor(7655)+sin(i + _j); end;
output;
drop i _j;
end;
run;
ods select none;
proc surveyselect data=original out=samp rep=10
sampsize=1000 method=srs;
run;
proc means data=samp;
by replicate;
var var1;
output out=_mean mean(var1)=mean1;
run;
data samp |
|
p*****o
发帖数: 543 | 26 数据如下：
VAR1
A
A
B
B
B
C
C
C
D
D
。。。。。
。。。。。
现在要做的就是给每个OBS都从1，2，3 （其实是1，2，3，。。。100，但是咱就先用3
个的来讨论吧）中随机选一个数，设为VAR2. 但是限制条件是：
VAR1 的值同样的情况下，VAR2的值必须不一样。
但是VAR1的值不同的情况下，VAR2的值可以一样。 |
|
p*****o
发帖数: 543 | 27 that's true for what i posted here....
but actually Im facing a more complicated situation. I have also other
variables, such as something like match_key. So once I need to keep track of
the original order....
Or probably I should change the way of posting the problem as we have
original data set:
Var1
A_1
A_2
B_1
B_2
B_3
C_1
C_2
C_3
...
and Im still trying to do the same thing as randomly pick 1 number from 1,2,
3 to assign to var1, while this can not be the same number for Var1 starti |
|
d*******o
发帖数: 493 | 28 data one ;
input Seller $ Year var1 var2 var3;
var=var1;output; var=var2;output; var=var3;output;
drop var1-var3;
cards;
ABC 2005 0 0 0
ABC 2006 4 0 0
ABC 2007 17 1 1
EFG 2005 32 4 3
EFG 2006 45 8 5
EFG 2007 50 3 12
;
run;
proc sort data=one; by seller; run;
proc transpose data=one out=two prefix=var; by seller; var var; run; |
|
b*****e
发帖数: 223 | 29 有没有一个 function 可以从一个 string 里面独独去掉我定义的单词
比如我有一个变量名 var1 用在 logistic model 里面，logistic model output 里的
effects 会有 var1 2:1, 我想要去掉 var1，只留 2:1 就好。有没有简单的
function 可以做？compress 是不行了 |
|
x*******u
发帖数: 500 | 30 my data:
var1 var2 var3 var4
2 4 6 7
4 9 7 6
5 2 1 1
如何得到一个新的variable， 它的值是var1-var4中有最大值的那个variable的名字。
结果应该是
newvar
var4
var2
var1
谢谢 |
|
e*******e
发帖数: 75 | 31 Hi,
I am using sql to create a new data set,
proc sql;
create table aaa as
select patid, max(var1) as var1, ....., max(var200) as var200
from bbb group by patid;
quit;
I am 200 variables var1 to var200 here. I am wondering if there is anyway
that I can avoid typing the 200 variables? Thank you very much. I would
appreciate the help. |
|
x*******u
发帖数: 500 | 32 从数据上看第一个变量的长度是10， 但是用你的code读出来结果是这样的：
char1 char2 char3
1 0 / 3 0 0 0 9 .1 4 1
中间还是有空格。
我用proc import读入数据后， log里面是这样的：
data WORK.READASC ;
%let _EFIERR_ = 0; /* set the ERROR detection macro variable */
infile 'myfile.csv' delimiter = ',' MISSOVER DSD lrecl=32767 firstobs=7 ;
informat VAR1 $21. ;
informat VAR2 $23. ;
informat VAR3 $9. ;
format VAR1 $21. ;
format VAR2 $23. ;
format VAR3 $9. ... 阅读全帖 |
|
g****8
发帖数: 2828 | 33 数据出来是这样的
id var1 var2
1 a1 b
1 a2 b
1 a3 b
2 a1 b
2 a2 b
2 a3 b
我想只用一个sql语句，加上两个count
结果是
id var1 var2 count（var1） count（id）
1 a1 b 3 2
1 a2 b 3 2
1 a3 b 3 2
2 a1 b 3 2
2 a2 b 3 2
2 a3 b 3 2
这样，我group by 的时候，应该group by哪个呀？如果group by var2的话，出来的... 阅读全帖 |
|
h********o
发帖数: 103 | 34 You can use array like this:
==================================
data test;
input var1 var3 var2 var10 var6;
array temp(5) var1 -- var6;
array logvar(5);
do i = 1 to dim(temp);
logvar(i) = log(temp(i));
end;
drop i var1 -- var6;
cards;
1 2 3 4 5
; |
|
t**j
发帖数: 418 | 35 data a;
set b;
by var1 var2;
if first.var1 or first.var2 then var4=var3;
retain var4; var4=var4*var3;run;
why for observations are first.var1 or first.var2, var4 is the square of
var3 but not var3?
thanks a lot! |
|
k*******a
发帖数: 772 | 36 data test;
input id seq var;
datalines;
1 1 11
1 4 22
1 5 33
1 6 44
1 7 55
1 8 66
2 1 11
2 2 22
2 8 33
3 1 11
3 2 22
3 3 33
3 4 44
3 5 55
3 6 66
3 7 77
3 8 88
;
run;
data test2;
set test(rename=(var=var1));
by id;
var = lag(var1);
if first.id then var=.;
drop var1;
run; |
|
e**h
发帖数: 46 | 37 data final;
merge one(where=(year=2004)
rename=(var1=var1_2004 var2=var2_2004 var3=var3_2004))
one(where=(year=2005)
rename=(var1=var1_2005 var2=var2_2005 var3=var3_2005))
one(where=(year=2006)
rename=(var1=var1_2006 var2=var2_2006 var3=var3_2006));
by 农场 地块;
run;
|
|
D*G
发帖数: 471 | 38 哇 我又来赚包子了。
前面那个是说var1和你的response variable完全相关。 Var1>0的时候response一直都
是0。 Google “complete separation”。 var1应该扔掉。
后面太乱了 没法看。 |
|
r*********8
发帖数: 4 | 39 我现在有2列char 的var, 我的目的是先把var中的每个“word" 分成一个个独立的
field, 然后看这两个var中有多少match的"word".
example:
原始数据：
var1 var2
looks like this looks like
disney disney cruise
----> 分解成：
var1_1 var1_2 var2_1 var2_2 var2_3
looks like this looks like
disney disney cruise
在var中每个“word"是用空格隔开的。我现在遇到的问题是，每个var含的"word”是不
一样的，可是我的做法好像就以最后一个obs的var的word个数来定loop的长度了。比如
上面的例子中，第一个obs，var1有2个word，第二个obs，var1只有一个word,结果do
loop的长度就成了1,然后就没有var1_2产生了。... 阅读全帖 |
|
H*******r
发帖数: 98 | 40 非高手，抛砖引玉
data newset;
set oldset(rename=(var=var1));
var= input(var1, best12.);
drop var1;
run;
位置问题，可以用attrib |
|
t*****w
发帖数: 254 | 41 * data three is what you wanted *;
data one;
input var1-var9;
cards;
1 2 3 7 9 9 4 7 10
1.1 4.1 4 3 3.4 7 7.4 8 7.3
2 2.1 2.5 2 2.1 5 2 2.1 2.5
;
data two three;
set one;
if var1 in (1, 4, 4,3 , 3.4, 7, 7.4, 8, 7.3,9) then output three;
else if 4 le var1 le 6 then output three;
array eight_var{8} var2-var9;
do i =1 to 8;
if eight_var{i} in (5, 5.1,5.5) then output three;
end;
output two;
run;
proc print data=three;
run; |
|
j******o
发帖数: 127 | 42 Try SQL procedure to count distinct var2 levels by var1;
proc sql;
create table check3 as
select var1, count(distinct var2) as levels
from have
group by var1;
quit; |
|
j******o
发帖数: 127 | 43 If you really want data step, may try below:
Sort your data HAVE by var1 var2 var3 first.
data obtain;
length lag_var3 $8;
set have;
by var1 var2 var3;
lag_var3=lag(var3);
if first.var2 then call missing(lag_var3, levels);
if var3^=lag_var3 then levels+1;
if last.var2;
keep var1 var2 levels;
run;
only |
|
j******o
发帖数: 127 | 44 data have;
input index var1 $ var1_description $ var2 $ var2_description $;
datalines;
1 cd02 cd02_text cd05 cd05_text
2 cd05 cd05_text cd03 cd03_text
3 cd03 cd03_text cd04 cd04_text
4 cd10 cd10_text cd08 cd08_text
;
run;
data obtain;
set have(keep=index var1 var1_description rename=(var1=var var1_
description=description))
have(keep=index var2 var2_description rename=(var2=var var2_
description=description));
by inde... 阅读全帖 |
|
d*****o
发帖数: 6 | 45 小女子在此跪谢大牛解答，
最后我的方法参考了Kirk君的scoring的方法，还是没骨气的用了sas，，掩面，，
data temp;
set test;
score=100*((var3='C')*3 + (var3='B')*2 + (var3 = 'A'))
+ 10*((var2='C')*3 + (var2='B')*2 + (var2 = 'A'))
+ ((var1='C')*3 + (var1='B')*2 + (var1 = 'A'))
proc sort data=temp;
by id descending score;
run;
data temp1;
set temp;
by id;
if first.id then output;
run;
Kirk君，求问max(calculated score) as maxscore是在每个id下的max score么。。
我试了你的code貌似不能啊。。还有having那部分不会变蓝log里报错。。。。。 |
|
k*******a
发帖数: 772 | 46 then you can use format to order the way you want
data test;
input id var1 $ var2 $ var3$;
datalines;
1 A B B
1 B C B
1 A B C
2 A A B
2 A A B
2 C C B
3 C A A
3 B A A
3 A A A
;
run;
proc format;
value $fmtest 'C' = 1
'B' = 2
'A' = 3;
run;
data test1;
set test;
var1f = put(var1, $fmtest.);
var2f = put(var2, $fmtest.);
var3f = put(var3, $fmtest.... 阅读全帖 |
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z*******e
发帖数: 32 | 47 大虾们：
对于下面var1和var2：
var1 = trim('Printing ')||put(750,3.);
var2 = trim('Printing ')||750;
为什么var1得到的是Printing750,而var2是Printing 750? |
|
j*****1
发帖数: 75 | 48 Sorry can't type Chinese.
Can anyone please help me with a SAS macro? I have a dataset includes
Month and Var1 (please see the dataset below) and I need to create a field
Var2, which is the mean of previous months' Var1.
For example, for Month 201401, var2= mean of 2.
for Month 201402 var2= mean of 2,4
for month 201404 var2 = mean of 2,4,5,8
etc......
Month Var1 Var2
201401 2 2.00
201402 4 3.00
201403 5 3.67
201404 8 4.75
201405 ... 阅读全帖 |
|
s*****r
发帖数: 223 | 49 刚学sas不久， 需要写macro 输出指定var 的最大值或最小值的obs（可能不只一个）。
exampledata
obs var1 var2
1 3 2
2 9 9
3 4 6
4 9 2
我知道可以
* sort to find the max or min;
proc sort data = exampledata out = sortedexampledata ;
by ascending/descending var1/var2;
run;
*determine how many obs to keep in the data set;
data sortedexampledata;
set sortedexampledata ( obs = N);
run;
现在的问题我不知道怎么coding 找到 这个 N， 我知道 N 就是 var1 or var 2 的
最大值或者最小值的 frequency， 怎么 输出这个frequency 然后传递给 N。
谢谢 |
|
s*****r
发帖数: 223 | 50 刚学sas不久， 需要写macro 输出指定var 的最大值或最小值的obs（可能不只一个）。
exampledata
obs var1 var2
1 3 2
2 9 9
3 4 6
4 9 2
我知道可以
* sort to find the max or min;
proc sort data = exampledata out = sortedexampledata ;
by ascending/descending var1/var2;
run;
*determine how many obs to keep in the data set;
data sortedexampledata;
set sortedexampledata ( obs = N);
run;
现在的问题我不知道怎么coding 找到 这个 N， 我知道 N 就是 var1 or var 2 的
最大值或者最小值的 frequency， 怎么 输出这个frequency 然后传递给 N。
谢谢 |
|
