Displaying 20 results from an estimated 7000 matches similar to: "merge for data.frame and matrix"
2010 May 11
1
create a data.frame for aov
Hi R-experts,
I try to find a way to transfer a matrix to a data.frame that is used as input of aov.
can you give me advice for that?
>mdat <- matrix(c(1,2,3, 11,12,13), nrow = 2, ncol=3, byrow=TRUE, dimnames = list(c("row1", "row2"), c("Col1", "Col2", "Col3")))
>mdat
Col1 Col2 Col3
row1 1 2 3
row2 11 12 13
===>
2010 Apr 29
2
by funtion
Hello,
I have a data.frame:
name col1 col2 col3 col4
AA 23 54 0.999 0.78
BB 123 5 1 0.99
AA 203 98 0.79 0.99
I want to get mean value data.frame in terms of name:
name col1 col2 col3 col4
AA 113.0000 76.0000 0.8945 0.8850
BB 123.00 5.00 1.00 0.99
I tried to use by function:
>aa<-by(test[,2:5], feature, mean)
2006 Mar 06
1
Sort problem in merge()
Hello!
I am merging two datasets and I have encountered a problem with sort.
Can someone please point me to my error. Here is the example.
## I have dataframes, first one with factor and second one with factor
## and integer
> tmp1 <- data.frame(col1 = factor(c("A", "A", "C", "C", "0", "0")))
> tmp2 <- data.frame(col1 =
2011 Mar 16
3
making dataframes
Dear all,
I have a dataframe which looks like this (dummy):
date<-c("jan", "feb", "mar", "apr", "may", "june", "july",
"aug","sep","oct","nov","dec")
col1<-c(8.2,5.4,4.3,4.1,3.1,2.5,1.1,4.5,3.2,1.9,7.8,6.5)
col2<-c(3.1,2.3,4.7,6.9,7.5,1.1,3.6,8.5,7.5,2.5,4.1,2.3)
2017 Aug 23
0
Comparing 2 dale columns
Patrick,
## Run the following script an notice the different values of the dataframe "data" in each instance.
# I understand you have done something like the following:
data <- data.frame(COL1 = c("6/1/14", "7/1/14"),
COL2 = c("5/1/15", "5/1/15"), stringsAsFactors = FALSE)
data$Date_Flag <- ifelse(data$COL2 >
2006 Mar 16
3
Did I use "step" function correctly? (Is R's step() function reliable?)
Hi all,
I put up an exhaustive model to use R's "step" function:
------------------------
mygam=gam(col1 ~ 1
+ col2 + col3 + col4
+ col2 ^ 2 + col3 ^ 2 + col4 ^ 2
+ col2 ^ 3 + col3 ^ 3 + col4 ^ 3
+ s(col2, 1) + s(col3, 1) + s(col4, 1)
+ s(col2, 2) + s(col3, 2) + s(col4, 2)
+ s(col2, 3) + s(col3, 3) + s(col4, 3)
+ s(col2, 4) + s(col3, 4) + s(col4, 4)
+ s(col2, 5) + s(col3,
2018 Feb 25
2
include
HI Jim and all,
I want to put one more condition. Include col2 and col3 if they are not
in col1.
Here is the data
mydat <- read.table(textConnection("Col1 Col2 col3
K2 X1 NA
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE)
The desired out put would be
Col1 Col2 col3
1 X1 0 0
2 K1 0 0
3 Y1 0 0
4 W1 0 0
6 K2 X1
2018 Feb 25
0
include
Hi Val,
My fault - I assumed that the NA would be first in the result produced
by "unique":
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE)
val23<-unique(unlist(mydat[,c("Col2","col3")]))
napos<-which(is.na(val23))
preval<-data.frame(Col1=val23[-napos],
2018 Feb 25
0
include
Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient?
--
Sent from my phone. Please excuse my brevity.
On February 25, 2018 7:55:55 AM PST, Val <valkremk at
2018 Feb 25
0
include
hi Val,
Your problem seems to be that the data are read in as a factor. The
simplest way I can think of to get around this is:
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE)
preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1],
Col2=NA,col3=NA)
rbind(preval,mydat)
2005 Oct 07
2
finding missing lines...
Take this as an example:
> a=data.frame(col1=c(1,2,3,4,5), col2=c
("my","beloved","daughter","son","wife"))
> b=data.frame(col1=c(1,2,4),
col2=c("my","beloved","son"))
> a
col1 col2
1 1 my
2
2 beloved
3 3 daughter
4 4 son
5 5 wife
> b
col1 col2
1 1 my
2
2018 Feb 25
2
include
Sorry , I hit the send key accidentally here is my complete message.
Thank you Jim and all, I got it.
I have one more question on the original question
What does this "[-1] " do?
preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1],
Col2=NA,col3=NA)
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2
2018 Feb 25
3
include
Thank you Jim,
I read the data as you suggested but I could not find K1 in col1.
rbind(preval,mydat) Col1 Col2 col3
1 <NA> <NA> <NA>
2 X1 <NA> <NA>
3 Y1 <NA> <NA>
4 K2 <NA> <NA>
5 W1 <NA> <NA>
6 Z1 K1 K2
7 Z2 <NA> <NA>
8 Z3 X1 <NA>
9 Z4 Y1 W1
On Sat, Feb 24, 2018 at 6:18 PM, Jim
2017 Aug 23
2
Comparing 2 dale columns
Thanks. But when I apply your codes I get all NA instead of TRUE and FALSE
________________________________
From: PIKAL Petr <petr.pikal at precheza.cz>
Sent: Wednesday, August 23, 2017 11:20:00 AM
To: Patrick Casimir; r-help at r-project.org
Subject: RE: Comparing 2 dale columns
Hi
your code is wrong.
I get
> test<-read.table("clipboard", header=T)
> str(test)
2018 Feb 24
0
include
Thank you Jim and all, I got it.
I have one more question on the original question
What does this "[-1] " do?
preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1],
Col2=NA,col3=NA)
mydat <- read.table(textConnection("Col1 Col2 col3
Z1 K1 K2
Z2 NA NA
Z3 X1 NA
Z4 Y1 W1"),header = TRUE)
2020 Sep 13
2
Cambiar datos de un DF
Hola:
Si codificas "col1" como un factor, lo puedes cambiar todo de una vez
renombrando los niveles correspondientes:
> df$col1<-as.factor(df$col1)
> df$col1
[1] uno dos 3 4 cinco 6 siete 8 9 diez
Levels: 3 4 6 8 9 cinco diez dos siete uno
> levels(df$col1)[1:5] <-
2017 Aug 23
0
Comparing 2 dale columns
Hi
your code is wrong.
I get
> test<-read.table("clipboard", header=T)
> str(test)
'data.frame': 2 obs. of 2 variables:
$ COL1: Factor w/ 2 levels "6/1/14","7/1/14": 1 2
$ COL2: Factor w/ 1 level "5/1/15": 1 1
> test$COL2<- as.Date(as.character(test$COL2, format="%y/%m/%d"))
> test$COL1<-
2008 Mar 16
2
How to loop through all the columns in dataframe
Hi:
Can anyone advice me on how to loop and perform a
calculation through all the columns.
here's my data
xd<-
c(2.2024,2.4216,1.4672,1.4817,1.4957,1.4431,1.5676)
pd<-
c(0.017046,0.018504,0.012157,0.012253,0.012348,0.011997,0.012825)
td<- c(160524,163565,143973,111956,89677,95269,81558)
mydf<-data.frame(xd,pd,td)
trans<-t(mydf)
trans
I have these values that I need to
2011 Jun 03
2
modify a data frame by values in the columns
I have a data frame like this:
col1 col2
r1 2 1
r2 4 3
r3 6 5
r4 8 7
r5 10 9
r6 12 11
r7 14 13
r8 16 15
r9 18 17
r10 20 19
I want to modify this data frame, for example, assign every row in column
col1 and col2 to -1 if the values in col1 is less than 12 and values in col2
is greater than 10. The result should look like this:
col1
2005 Oct 09
3
[ subscripting sometimes loses names (PR#8192)
--rwEMma7ioTxnRzrJ
Content-Type: text/plain; charset=us-ascii
Content-Disposition: inline
R, like recent versions of S-Plus, sometimes - but not always - loses
names when subscripting objects with "[". (Earlier versions of S and
S-Plus had the correct, name-preserving behavior.) This seems bad, it
would be better to remove names only by explicit request, not as an
accidental