Displaying 20 results from an estimated 1000 matches similar to: "rows process in DF"
2010 Jul 27
4
re-sampling of large sacle data
myDF:
d1 d2 d3 d4 d5
-0.166910351 0.022304377 -0.00825924 0.008330689 -0.000925938
-0.166910351 0.022304377 -0.00825924 0.008330689 -0.000925938
-0.166910351 0.022304377 -0.00825924 0.008330689 -0.168225938
-0.166910351 0.022304377 -0.00825924 0.008330689 -0.168225938
-0.166910351 0.022304377 -0.00825924 0.008330689 -0.168225938
-0.166910351
2004 Nov 20
2
subset on data frame
I have a data frame. And I'd like to subset according to rownames.
subset(mydataframe, rownames(mydataframe) == myrow, select = mycols)
it turned out that "myrow" cannot be a vector. But I have multiple rows to
pick. Is there a way to get around this problem??
Thank you for your help!!
Lei Jiang
Department of Chemsitry
University of Washington
Box 351700
Seattle, WA 98195
2010 Jul 19
3
invalid type error
>myDF =
data.frame(id=c("A10","A20"),d1=c(.3,.3),d2=c(.4,.4),d3=c(-.2,.5),d4=c(-.3,.6),d5=c(.5,-.2),d6=c(.6,-.4),d7=c(-.9,-.5),d8=c(-.8,-.6))
>doit=function(x)c(x[1],sum_LK_positive=sum(x[-1][x[-1]>0]),sum_LK_negative=sum(x[-1][x[-1]<0]))
> myDF
id d1 d2 d3 d4 d5 d6 d7 d8
1 A10 0.3 0.4 -0.2 -0.3 0.5 0.6 -0.9 -0.8
2 A20 0.3 0.4 0.5 0.6 -0.2
2007 Feb 09
3
How to count the number of NAs in each column of a df?
I would like to remove columns of a df which have too many NAs.
I think that summary() should give me the information, I just don't
know how to access it.
Advice?
_____________________________
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400 Charlottesville, VA 22904-4400
Parcels: Room 102 Gilmer Hall
McCormick Road
2004 Jul 29
3
Editing Strings in R
I was wondering if there is a way of editting strings in R. I
have a set of strings and each set is a row of numbers and paranthesis.
For example the first row is:
(0 2)(3 4)(7 9)(5 9)(1 5)
and I have a thousand or so such rows. I was wondering how I
could get the corresponding string obtained by adding 1 to all the
numbers in the string above.
Dursun
[[alternative HTML version deleted]]
2013 Feb 03
1
Looping through rows of all elements of a list that has variable length
Dear R-ers,
I have a list of data frames such that the length of the list is unknown in
advance (it could be 1 or 2 or more). Each element of the list contains a
data frame.
I need to loop through all rows of the list element 1 AND (if applicable)
of the list element 2 etc. and do something at each iteration.
I am trying to figure out how to write a code that is generic, i.e., loops
through the
2017 Nov 22
6
assign NA to rows by test on multiple columns of a data frame
Given this data frame (a simplified, essential reproducible example)
A<-c(8,7,10,1,5)
A_flag<-c(10,0,1,0,2)
B<-c(5,6,2,1,0)
B_flag<-c(12,9,0,5,0)
mydf<-data.frame(A, A_flag, B, B_flag)
# this is my initial df
mydf
I want to get to this final situation
i<-which(mydf$A_flag==0)
mydf$A[i]<-NA
ii<-which(mydf$B_flag==0)
mydf$B[ii]<-NA
2009 Jan 20
5
Problem with subset() function?
Hi all,
Can anyone explain why the following use of
the subset() function produces a different
outcome than the use of the "[" extractor?
The subset() function as used in
density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age)))
appears to me from documentation to be equivalent to
density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])
2011 May 19
1
Creating a "shifted" month (one that starts not on the first of each month but on another date)
Hello!
I have a data frame with dates. I need to create a new "month" that
starts on the 20th of each month - because I'll need to aggregate my
data later by that "shifted" month.
I wrote the code below and it works. However, I was wondering if there
is some ready-made function in some package - that makes it
easier/more elegant?
Thanks a lot!
# Example data:
2005 Dec 08
3
Reshaping data
Dear all,
given I have data in a data.frame which indicate the number of people in
a
specific year at a specific age:
n <- 10
mydf <- data.frame(yr=sample(1:10, size=n, replace=FALSE),
age=sample(1:12, size=n, replace=FALSE),
no=sample(1:10, size=n, replace=FALSE))
Now I would like to make a matrix with (in this simple example)
10 columns (for the
2005 Feb 03
2
Surprising Behavior of 'tapply'
Dear all,
I wanted to make a two-way-table of two variables with a counting
variable stored in another column of a dataframe. In version 1.9.1, the
behavior is as expected as shown in the simplified example code.
> sex <- rep(c("F", "M"), 5)
> income <- c(rep("low", 5), rep("high", 5))
> count <- 1:10
> mydf <-
2007 Sep 01
2
Comparing "transform" to "with"
Hi All,
I've been successfully using the with function for analyses and the
transform function for multiple transformations. Then I thought, why not
use "with" for both? I ran into problems & couldn't figure them out from
help files or books. So I created a simplified version of what I'm
doing:
rm( list=ls() )
x1<-c(1,3,3)
x2<-c(3,2,1)
x3<-c(2,5,2)
2001 Nov 05
3
vector problems
I dont get it:
> is.vector(c(mydf[1]))
[1] TRUE
> unique(c(mydf[1]))
Error in unique(c(mydf[1])) : unique() applies only to vectors
>
Is it a vector or not? This stuff is driving me nuts. I'm simply trying
to convince R that my grouping vector is actually a vector so that
unique will work. Its just a vector of numbers, so why shouldnt it work?
--
2013 Feb 26
2
merging or joining 2 dataframes: merge, rbind.fill, etc.?
#I want to "merge" or "join" 2 dataframes (df1 & df2) into a 3rd
(mydf). I want the 3rd dataframe to contain 1 row for each row in df1
& df2, and all the columns in both df1 & df2. The solution should
"work" even if the 2 dataframes are identical, and even if the 2
dataframes do not have the same column names. The rbind.fill function
seems to work. For
2017 Nov 22
1
assign NA to rows by test on multiple columns of a data frame
...well, I don't think this is exactly the expected result (see my post)
to be noted that the columns affected should be "A" and "B"
thanks for the help
max
----- Messaggio originale -----
Da: "Rui Barradas" <ruipbarradas at sapo.pt>
A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it>, "r-help" <r-help at
2008 Jan 03
2
retaining formatting when converting a vector to a matrix/data.frame?
Please see example code below.
I have a vector ("mydata") of length 10. "mydata" can have various formats (e.g. numeric, text, POSIXct, etc) I use the matrix and data.frame functions to convert "mydata" to a dataframe ("mydf") of 2 columns and 5 rows.
What is a "good" way to ensure that the format is retained when I create the
2007 Sep 02
2
NAs in indices
Hi All,
I'm fiddling with an program to read a text file containing periods that
SAS uses for missing values. I know that if I had the original SAS data
set instead of a text file, R would handle this conversion for me.
Data frames do not allow missing values in their indices but vectors do.
Why is that? A search of the error message points out the problem and
solution but not why they
2011 Jun 09
1
Error: missing values where TRUE/FALSE needed
I'm writing a function and keep getting the following error message.
myfunc <- function(lst) {
lst <- list(roots = c("car insurance", "auto insurance"),
roots2 = c("insurance"), prefix = c("cheap", "budget"),
prefix2 = c("low cost"), suffix = c("quote", "quotes"),
suffix2 = c("rate",
2013 Feb 18
1
ggplot2 and facet_wrap help
Dear R experts,
I am trying to arrange multiple plots, creating one graph for each
size1 factor variable in my data frame, and each plot has the median
price on the y-axis and the size2 on the x-axis grouped by clarity:
library(ggplot2)
df <- data.frame(price=matrix(sample(1:1000, 100, replace = TRUE), ncol = 1))
df$size1 = 1:nrow(df)
df$size1 = cut(df$size1, breaks=11)
2011 Aug 16
1
Utilizing column names to multiply over all columns
## Hello there,
## I have an issue where I need to use the value of column names to
multiply with the individual values in a column and I have many
columns to do this over. I have data like this where the column names
are numbers:
mydf <- data.frame(`2.72`=runif(20, 0, 125),
`3.2`=runif(20, 50, 75),
`3.78`=runif(20, 0, 100),
yy=