t********m
发帖数: 939 | 1 【 以下文字转载自 DataSciences 讨论区 】
发信人: tulipdream (xiaohuaidan), 信区: DataSciences
标 题: 请教一个R问题：怎么rbind一系列data，如data1，data2,....data1000
发信站: BBS 未名空间站 (Thu Nov 13 17:35:36 2014, 美东)
有没有一个简单点的方法，类似于rbind(data1-data1000)的，求各位不吝赐教，谢谢
！ |
|
d********t
发帖数: 837 | 2 Reduce(rbind, list(data1,data2,data3,data4))
example:
Reduce(rbind, list(data.frame(x1=c(1,2,3),x2=c(2,3,4)),data.frame(x1=c(5,6,7
),x2=c(7,5,4)),data.frame(x1=c(5,4,3),x2=c(7,6,5)))) |
|
s******e
发帖数: 2181 | 3 谢谢解惑，我的问题是这样的，这是一个被我简化后的在matlab环境下运行的C程序。
我想测试我输入的数据是否被正确读入。一个途径是通过把输入参数赋给输出的地址，
来得到结果，看来是正确的。另一个是通过printf把输入数据直接打出来，可是这个打
出来的数字始终为0，无论采用printf("input=%dn", data2[0]); 还是printf("input=
%dn", *data2);都一样。真是见鬼了，那data2这个地址上存的数据到底是什么
#include "mex.h"
#include "gpu/mxGPUArray.h"
#include "cuda.h"
void mexFunction(int nlhs, mxArray *plhs[],
int nrhs, mxArray const *prhs[])
{
double *data1; *data2;
int m,n;
m=mxGetM(prhs[0]);
n=mxGetN(prhs[0]);
plhs[0]=mxCreateDoubl... 阅读全帖 |
|
t**********r
发帖数: 182 | 4 Want to merge two data sets using proc sql:
Data1:
var1 var2 date1
Data2:
var1 var2 date2 rating
(Note: var1 and var2 are the same variables in these two data sets)
Question:
Select rating in data2 to data1; meeting the following criteria:
1. date1 - date2 >0
2. date1 - date2 has the minimum value.
I wrote the following code; but it won't work:
proc sql;
create table data3 as
select data1.*, data2.rating, date1-date1 as diff
from data1, data2
where data1.var1=data2.var1 and data1.var2=data2.var2 |
|
w****g
发帖数: 597 | 5 Thank you for your reply.
Yes. I already boot up to a running system with the live cd. I saw the
partitions in file manager with names, for example, DATA2. (please see photo
proof at 2nd reply, the left bottom corner of that attached photo show '
DATA2')
Then, I had my question: Is DATA2 partition /dev/sdb1 or sdb2, or
sdb3,..., or sdb5, how can I find the information?
Here is reason: I am planning to install Ubuntu into my DATA2 partition at external USB hard drive because DATA2 is blank. But, |
|
t**********r
发帖数: 182 | 6 Want to merge two data sets using proc sql:
Data1:
var1 var2 date1
Data2:
var1 var2 date2 rating
(Note: var1 and var2 are the same variables in these two data sets)
Question:
Select rating in data2 to data1; meeting the following criteria:
1. date1 - date2 >0
2. date1 - date2 has the minimum value.
I wrote the following code; but it won't work:
proc sql;
create table data3 as
select data1.*, data2.rating, date1-date1 as diff
from data1, data2
where data1.var1=data2.var1 a |
|
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 |
|
s*****n
发帖数: 2174 | 8 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 |
|
t**********r
发帖数: 182 | 9 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 | 10 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; |
|
p******p
发帖数: 13 | 11 亲测可用，想覆盖data1的话把最后的new_data1改成data1就好，虽然覆盖源dataset习
惯很不好。
data data1;
input id t1 t2;
datalines;
1 2 3
2 4 5
3 4 5
;
run;
data data2;
input id;
datalines;
1
2
;
run;
proc sql noprint;
create table new_data1 as
select data1.*,coalesce(flag,0) as flag from
data1 left join (select data2.*,1 as flag from data2)
on data1.id=data2.id
;
quit; |
|
s*******s
发帖数: 1031 | 12 follow一下我的面经。
http://www.mitbbs.com/article_t/JobHunting/32517841.html
整理了我的几个解答的算法，分享一下。欢迎批评指正。
多谢！
1. 写一个程序，找出 5^1234566789893943的从底位开始的1000位数字。
我用的递归+数组大数乘法。
// Caclulate (m^n)%(10^k). Keep the k integer numbers in an array.
// Note: the integer numbers are in reversed in the array
// Assume: m>0, n>0, k>0
// Need to check validity outside of this function.
// call calculate(5, 1234566789893943, 1000) to get result.
// Time complexity: O((log n) * k * k)
// Space complexity: O((log n) * k)
ve... 阅读全帖 |
|
s*******s
发帖数: 1031 | 13 follow一下我的面经。
http://www.mitbbs.com/article_t/JobHunting/32517841.html
整理了我的几个解答的算法，分享一下。欢迎批评指正。
多谢！
1. 写一个程序，找出 5^1234566789893943的从底位开始的1000位数字。
我用的递归+数组大数乘法。
// Caclulate (m^n)%(10^k). Keep the k integer numbers in an array.
// Note: the integer numbers are in reversed in the array
// Assume: m>0, n>0, k>0
// Need to check validity outside of this function.
// call calculate(5, 1234566789893943, 1000) to get result.
// Time complexity: O((log n) * k * k)
// Space complexity: O((log n) * k)
ve... 阅读全帖 |
|
y***n
发帖数: 1594 | 14 原题在这里。 http://www.mitbbs.com/article_t/JobHunting/32517841.html 但是觉得不太对
搞这些题觉得很没思路。
// Caclulate (m^n)%(10^k). Keep the k integer numbers in an array.
// Note: the integer numbers are in reversed in the array
// Assume: m>0, n>0, k>0
// Need to check validity outside of this function.
// call calculate(5, 1234566789893943, 1000) to get result.
// Time complexity: O((log n) * k * k)
// Space complexity: O((log n) * k)
vector calculate(unsigned long m, unsigned long n, int k) {
if(... 阅读全帖 |
|
m**********2
发帖数: 2252 | 15 有这样一个table，想直接select出不同ID的data1,data2。。。的sum，该怎莫做？
原来的：
ID Data1 Data2 Data3
1 3341 1926 336
1 972 562 130
1 1580 760 213
1 742 345 87
1 7776 4237 889
2 27129 15347 3002
2 1428 825 139
2 3211 1918 361
2 1114 598 114
2 1219 602 110
3 1238 684 127
select 之后
ID Data1 Data2 Data3
1 14411 7830 1655
2 34101 19290 3726
3 1238 684 127
谢！！！ |
|
t***5
发帖数: 832 | 16 【 以下文字转载自 Linux 讨论区 】
发信人: take5 (NotAvailable), 信区: Linux
标 题: 怎样mount另一个Linux server的filesystem
发信站: BBS 未名空间站 (Mon Jan 7 12:30:10 2013, 美东)
Two linux servers both running Red Hat 6.
One (linux1) has /dev/sdb1 mounted on /data1, another (linux2) has /dev/sdb2
mounted on /data2. How to make two filesystems available to both servers?
tried something like
http://etutorials.org/Linux+systems/red+hat+linux+bible+fedora+
Thought this should be a simple task. But using the command on linux1
mount l... 阅读全帖 |
|
N****w
发帖数: 21578 | 17 right click on DATA2, check its property.
if still not see, run command
df
or
cat /etc/mtab :)
photo
external USB hard drive because DATA2 is blank. But, Ubuntu Installer step_
4 shows only /sdb1, sdb5,sdb3,sdb4, and it did not show which sdb partition
is DATA2 (please see att |
|
t***5
发帖数: 832 | 18 Two linux servers both running Red Hat 6.
One (linux1) has /dev/sdb1 mounted on /data1, another (linux2) has /dev/sdb2
mounted on /data2. How to make two filesystems available to both servers?
tried something like
http://etutorials.org/Linux+systems/red+hat+linux+bible+fedora+
Thought this should be a simple task. But using the command on linux1
mount linux2:/data2 /data2
didn't work. Any idea?
Thanks! |
|
z****u
发帖数: 23 | 19 在matlab中如何运行不在current working directory中的script?
比方说现在的working directory是"C:\Program\Files\MATLAB\R2007b\work", 在\
work下面有一些文件夹，比方说有\work\data1,\work\data2等等，里面都有一些
script，比方是\work\data1\code1, \work\data2\code2。能在\work里写个script，
运行\work\data1\code1, \work\data2\code2吗？
多谢多谢！ |
|
y****2
发帖数: 34 | 20 data <- matrix(c(1 ,1 ,2, 2, 1, 3, 4, 2,1, 5, 6, 3,2, 7, 8, 3,2, 9, 10, 4),
ncol=4, byrow=T)
colnames(data) <- c("id", "x1", "x2", "e")
### step1:
data[,2:3] <- data[,2:3]*data[,4]
data1 <- data[,1:3]
### step2:
data2 <- aggregate(data1[,2:3], list(id=data1[,1]), sum)
### step3:
data3 <- split(data2[,2:3], f=list(data2[,1]))
data3 <- lapply(data3, as.vector, "numeric")
mprod <- function(x){x %*% t(x)}
data4 <- lapply(data3, mprod)
### step4:
data5 <- 0
for(i in 1:length(data4)){
data5 <- data5 + |
|
k*****u
发帖数: 1688 | 21 正解
昨天网上刚刚看了
union合并了,data1的变量比data2多,那么变量名都是data1的
要是data2的变量比data1多,那么前面的用data1变量名,后面用data2变量名 |
|
s*****n
发帖数: 2174 | 22 其实就是一个循环, 循环里面包含一个判断. 实现的话在R里也就十几行.
data <- read.table(...)
result <- data.frame(try = 1:1000, output = NA, case = NA)
for (i in 1:1000){
data1 <- data[sample(100000, 10000), ]
data2 <- data[sample(100000, 10000), ]
if (mean(data1$var1) > 0){
fit1 <- lm(...)
result$output[i] <- functionA(data2, fit1$parameter_a)
result$case[i] <- "A"
} else {
fit2 <- glm(...)
result$output[i] <- functionB(data2, fit2$parameter_b)
result$case[i] <- "B"
}
}
hist(result$output[ |
|
w********5
发帖数: 72 | 23 This is my answer. My codes are alway very long and not efficient. Please
help simlify.
data data1;
input var1;
cards;
5
6
;
run;
data data2;
input var2;
cards;
5
6
;
run;
data new;
infile datalines dlm=" ";
input name $ var $ ;
datalines;
data1 var1
data2 var2
data2 var2
data4 var4
;
run;
proc sql;
select name into:name1-:name&SYSMAXLONG
from new;
select var into:col1-:col&&SYSMAXLONG
from new;
quit;
%put _user_;
option mprint mlogic;
%macro mutiple;
%do i=1 %to &sqlobs;
proc so |
|
e****t
发帖数: 766 | 24 a lazy way to Q1: you can try to see if it works.
for data2, you have each X vector as a record, set all the y values to be .
, set to data1, fit the model, since all y in data2 is missing, it wont
participate the estimation, but when you specify 'pred', the missing y will
be predicted.
for Q2, if it were me, i will either delete X or only select X with CA,OR，
WA from data2 to calculate. |
|
z***9
发帖数: 1052 | 25 我有一个data1
ID Description
1 aadd
2 adsd
3 asdd
....
现在我又有了一个data2
ID Description
1 aaddd
2 adsdq
3 asddg
4 fdsfg
....
我希望用data2里的Description来替换data1里的Description,在ID相等的情况下.
我想到的笨办法就是做个left join,新建一个table.有没有什么fancy的方法直接
update data1 里的Description到data2的Description. |
|
s*****n
发帖数: 2174 | 26 R当初是为了run一些统计的东西设计的, 主要考虑的是方便, 所以很多object
handling都过于flexible以至于牺牲了performance. 举个最简单的例子, 就说data
frame吧, 其实就是个很慢的东西. 很多时候data manipulation的时候, 用matrix会快
的多.
data1 <- matrix(rep(0, 1e6), ncol = 1000, nrow = 1000)
data2 <- data.frame(data1)
object.size(data1)
object.size(data2) # almost same size
# 0.42秒
for (i in 1:1000){
for (j in 1:1000){
data1[i,j]
}
}
# 33.2秒
for (i in 1:1000){
for (j in 1:1000){
data2[i,j]
}
}
这几乎是100倍的速度差别, 就仅仅是data frame vs matrix. 两个例子里面都有双层
for loop. 其实fo... 阅读全帖 |
|
|
c**y
发帖数: 419 | 28 MultiCharts可以引用2个股票的数据, 所以可以实现简单的2个股票的long/short交易.
如果第二个股票用大盘指数ETF或者行业ETF的话, 就变成了alpha/beta分离交易.
引用方法是建立指标公式 spread=close of data1 - beta * close of data2 ; 这样
这个spread就可以被图形化监控了.
由于有下面这个近似公式: (1+x)/(1+y)=1+(x-y)
x, y 分别是股票1,2的期间return%
所以一般beta取1的话,我们就直接用spread=close of data1 / close of data2. 这样
比较方便.
下图是AAPL(图1) 对RSP(等权重SP500指数ETF, 图2)的long/short对的比率(图3中红色
), 和2%的跟踪止损(图3中黄色). |
|
r****y
发帖数: 26819 | 29 這樣的效果很常見
比如6行javascript代碼就能讓safari 4.0.4 死掉：
It |
|
c******n
发帖数: 4965 | 30 【 以下文字转载自 Linux 讨论区 】
【 原文由 creation 所发表 】
input trash.mp
beginfig(1)
draw begingrf(5in,4in);
rdata("datafile",1);
rdata("datafile",2);
rdata("data2",1);
rdata("data2",2);
matplot("0 1 cg ddash t*");
matplot("2 3 cr ddot tx");
ylabel("blahblah");
xlabel("red is sqrt(x)*1000, green is sqr(x)");
finishgrf; endfig;
end |
|
l*******c
发帖数: 523 | 31 如果有个t的函数：
V(t) = sqrt { (Va)^2 + [(Vb)^2 - (Va)^2]*[1-e^(kt/20)]/(1-e^T)};
Va是已知的值；
Vb也是已知的值；
k, T是常数。
然后k从1到20，有没有简单的算法可以减少for循环里运算量？以下是那段code：
if (Vb > Va)
deltaDAC = (Vb * Vb - Va * Va)/(1-exp(T));
else
deltaDAC = (Va * Va - Vb * Vb)/(1-exp(T));
if (Vb > Va)
{
for (k=1; k <= 20; k++)
{
Data1 = Va * Va;
Data2 = deltaDAC * (1-exp(k/20));
DAC_Set_Value = sqrt(Data1 + Data2);
DAC_Outp... 阅读全帖 |
|
t***5
发帖数: 832 | 32 是不是作以下就可以了?
on server side:
#service nfs start
on client side:
#mount -t nfs linux2:/data2 /data2
但总是得到
mount.nfs: Connection timed out
请问这是什么问题? |
|
w****g
发帖数: 597 | 33 附上问题屏幕贴图 ：
1，在右上角的窗口是install“step 4 of 7 ”的屏幕;可以看到的选项都是/sda内置盘的分区,而没有/sdb 外置USB盘的任何分区. 如果要在外置USB盘/sdb在安装，哪里可以选择外置USB盘/sdb？
2，在左下角的terminal是disk information 'fdisk -l', 和桌面背景上的外置USB盘的4个分区sdb1,sdb3,sdb4,sdb5. 这里4个分区都已经NTFS格式化,其卷标名字分别是"Charlie","DATA1","DATA2","DATA3". 在分区/dev/sdb1,2,3,4,5之中，哪一个是安装的目标盘"DATA2"?
谢谢． |
|
N****w
发帖数: 21578 | 34 boot up with the live CD, plugin your USB HD
run a file manager program, the 4 partitions will be loaded automatically
and you will see the "name" ...
置盘的分区,而没有/sdb 外置USB盘的任何分区. 如果要在外置USB盘/sdb在安装，哪里
可以选择外置USB盘/sdb？
盘的4个分区sdb1,sdb3,sdb4,sdb5. 这里4个分区都已经NTFS格式化,其卷标名字分别是
"Charlie","DATA1","DATA2","DATA3". 在分区/dev/sdb1,2,3,4,5之中，哪一个是安
装的目标盘"DATA2"? |
|
w****g
发帖数: 597 | 35 Thank you for your reply.
I already found the external USB hard drive at Ubuntu installer step 4
successfully. There are partition /dev/sdb1,5,3,and 4. (please see my
following new post).
The next step is identify which NTFS partition of 4 partitions is the blank
NTFS partition labeled with "DATA2"? or which partition, sdb1, 5,3,or 4 is
the blank NTFS partition "DATA2", so I can install Ubuntu ?
http://www.mitbbs.com/article_t/Linux/31189560.html
Thank you in advance. |
|
l*******c
发帖数: 523 | 36 如果有个t的函数：
V(t) = sqrt { (Va)^2 + [(Vb)^2 - (Va)^2]*[1-e^(kt/20)]/(1-e^T)};
Va是已知的值；
Vb也是已知的值；
k, T是常数。
然后k从1到20，有没有简单的算法可以减少for循环里运算量？以下是那段code：
if (Vb > Va)
deltaDAC = (Vb * Vb - Va * Va)/(1-exp(T));
else
deltaDAC = (Va * Va - Vb * Vb)/(1-exp(T));
if (Vb > Va)
{
for (k=1; k <= 20; k++)
{
Data1 = Va * Va;
Data2 = deltaDAC * (1-exp(k/20));
DAC_Set_Value = sqrt(Data1 + Data2);
DAC_Outp... 阅读全帖 |
|
V********n
发帖数: 3061 | 37 那你把下面三步合为一步，可以减少两次生成中间变量并寻址的动作，对减少时间有帮
助：
Data1 = Va * Va;
Data2 = deltaDAC * (1-exp(k/20));
DAC_Set_Value = sqrt(Data1 + Data2);
改成：
DAC_Set_Value = sqrt(Va * Va + deltaDAC * (1-exp(k/20)));
说实在的，现在的编程上很少会需要去考虑这么细微的区别。如果你真的对时间抠到纳
秒的地步，这么做也许会有点帮助。
ringing. |
|
l*******c
发帖数: 523 | 38 如果有个t的函数：
V(t) = sqrt { (Va)^2 + [(Vb)^2 - (Va)^2]*[1-e^(kt/20)]/(1-e^T)};
Va是已知的值；
Vb也是已知的值；
T是常数。
然后k从1到20，有没有简单的算法可以减少for循环里运算量？以下是那段code：
if (Vb > Va)
deltaDAC = (Vb * Vb - Va * Va)/(1-exp(T));
else
deltaDAC = (Va * Va - Vb * Vb)/(1-exp(T));
if (Vb > Va)
{
for (k=1; k <= 20; k++)
{
Data1 = Va * Va;
Data2 = deltaDAC * (1-exp(k/20));
DAC_Set_Value = sqrt(Data1 + Data2);
DAC_Output(... 阅读全帖 |
|
s*****n
发帖数: 2174 | 39 1. 用 is.element 很容易搞定.
data1[!is.element(paste(data1$key1, data1$key2), paste(data2$key1, data2$
key2)), ]
2. 直接tapply就好了, 根本用不着什么new啊, index的.
tapply(data$z, list(data$x, data$y), sum) |
|
q**j
发帖数: 10612 | 40 look up PROC TRANSPOSE. I strongly suspect this one can give you the shortes
t, maybe, fastest codes. and you do not need data one at all.
proc transpose data = data2 out = data2;
by id;
run;
you might or might not need other stuff like 'id' statement. |
|
g*******y
发帖数: 380 | 41 说实在的,不懂你在说什么?
第一,你所指的保留值是什么?是指你在data1里提取的count的值为最大的组里对应的
num1,num2吗?
第二,不懂你为什么要想那么多复杂的方法?因为从你的sample来看,你的每个组只是根
据count的值来进行重复,count是多少,每个组重复的观察值就是几个.你如果只是想要
count最大组里的num1,num2来剔取data2里的观察值的话.我觉得最简单的方法:
proc sort data=data1; by descending count;
用最大count组里的值,比如x,y来剔除你不要的观察值.
data temp; set data2; if num1=x and num2=y;
如果这个不是你想要的,那么也许我没有看懂你的问题. |
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a****m
发帖数: 693 | 42 谢谢，主要是行数太多，前62个数（row wise）是一组，如果matrix 置换会把所有的
行弄混，结果txt转换成cvs时候，把空格给变成0了，用genelowvalfilter 好像也不能
去掉，只好这样处理了：
谢谢
[n,p]=size(data);
for i=1:2000
gene(i,:)=reshape(data(((i-1)*16+1):16*i,1:4),1,64);
end
data1=gene(:,1:47);
data2=gene(:,49:63);
data=[data1,data2];
这个filter 不起作用。
[mask,genes] = genelowvalfilter(gene,'absval',1); |
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i**f
发帖数: 1195 | 43 不用看文字，看data就好了
把data1的格式变成data2，加上ADDL和II（additional dose的数量和dosing interval）
data1
D Date Time Event Glucose
101 01Jan2000 7:30 Sampling .
101 01Jan2000 8:00 Dosing .
101 01Jan2000 8:30 Sampling 100
101 01Jan2000 9:30 Sampling 200
101 01Jan2000 12:30 Sampling 170
101 01Jan2000 18:30 Sampling 140
101 02Jan2000 7:30 Sampling 90
101 02Jan2000 8:00 Dosing .
101 03Jan2000 8:00 Dosing .
101 04Jan2000 8:00 Dosing .
101 05Jan2000 8:00 Dosing .
data2 |
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s********p
发帖数: 637 | 44 She is pulling data from large tables, in most case, she will create new
dataset containing pulled variable for further analysis.
I am not clear if temporary will be created when no new dataset needed. I
don't use system option(fullstimer), but just tried the following,
proc sql;
select p1.id from
data1 p1
inner join
data2 p2 on
p1.id=p2.id
;
quit;
and check if temporary files generated and found "#tf0005.sas7butl" created and size changed.
ll /saswork/SAS_workDD3300003446
total 7000
-rw-rw-r-- ... 阅读全帖 |
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c*****1
发帖数: 131 | 45 I think oloolo give the best answer for Q2.
However, in actual business, this can happen from time to time. We can not consult back with business and data side always. There should be a way to handle this kind of situation.
Add my 2 cents:
Check the distribution of variable x in data1 and data2. If frequency of x in ('TX' and 'NV') are small, and there are other predictors in the model besides x, you can still fit the model if you create indicators like this (it will take 'TX' and 'NV' into acco... 阅读全帖 |
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t**c
发帖数: 539 | 46 Using MODIFY statement.
Let data1 be the master data set and data2 be the transaction-data-set:
DATA data1;
Modify data1 data2;
By ID;
RUN; |
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p***r
发帖数: 920 | 47 or you can do it in another brutal way
*WIDE TO LONG;
PROC TRANSPOSE DATA=data1 OUT=data2;
BY var_id;
VAR _ALL_;
RUN;
data data3;
set data2;
log_var=log(col1);
run;
*LONG TO WIDE;
PROC TRANSPOSE DATA=data3 OUT=data4
BY var_ID;
ID variable;
VAR log_var;
RUN; |
|
f******u
发帖数: 250 | 48 you do not need a macro.
data data1(keep=x);
input variableName $14. variableLength : $12. price;
x=catx(' ', variableName,variableLength,price);
datalines;
FundingAmount 0-15000 0.217
ObjectAmount 15000-30000 0.318
FundingAmount 30000-60000 0.519
;
run;
data data2(keep=x);
input firstname : $7.lastname : $7. age;
x=catx(' ',firstname,lastname,age);
datalines;
Ashley Liu 17
brandon green 30
susan Chen 28
run;
data data3;
set data1 data2;
run;
data _null_;
fil... 阅读全帖 |
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