Displaying 20 results from an estimated 5000 matches similar to: "Converting list of lists into dataframes"
2018 May 10
2
using for loop with data frames.
Hi,
Is it possible use a loop to process many data frames in the same way?
For example, if I have three data frames, all with same variables
df_bs_id1 <- read.csv("test1.csv",header =TRUE)
df_bs_id2 <- read.csv("test2.csv",header =TRUE)
df_bs_id3 <- read.csv("test3.csv",header =TRUE)
How could I would implement a code loop that , for instance, would
2011 Apr 13
2
setting pairwise comparisons of columns
Hi,
I have a number of genes (columns) for which I want to examine pairwise
associations of genotypes (each row is an individual)...For example (see
data below), I would like to compare M1 to M2, M2 to M3, and M1 to M3 (i.e.
does ac from M1 tend to be found with bc from M2 more often than expected.)
Down stream I will be performing chi square tests for each pair.
But I am looking for a way to
2009 Dec 03
2
Dataframe help
Hi there
I have two dataframes
Dataframe_1
column_1 colum_2
121 12345
145 1675
167 2765
Dataframe_2
column_1 column2
121 abc
345 lmn
167 efg
I want a resulting dataframe
121 12345 abc
167 2765 efg
how do i go abt it
Ramya
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Sent from the R
2010 May 31
2
accessing a data frame with row names
Readers,
I have entered a file into r:
,column1,column2
row1,0.1,0.2
row2,0.3,0.4
using the command:
dataframe<-read.table("/path/to/file.csv",header=T,row.names=1)
When I try the command:
dataframe[,2]
I receive the response:
NULL
I was expecting:
row1 0.2
row2 0.4
What is my error with the syntax please?
Yours,
r251
mandriva2009
2004 Nov 23
5
number of pairwise present data in matrix with missings
is there a smart way of determining the number of pairwise present data
in a data matrix with missings (maybe as a by-product of some
statistical function?)
so far, i used several loops like:
for (column1 in 1:99) {
for (column2 in 2:100) {
for (row in 1:500) {
if (!is.na(matrix[row,column1]) & !is.na(matrix[row,column2])) {
pairs[col1,col2] <- pairs[col1,col2]+1
2009 May 14
2
Function to read a string as the variables as opposed to taking the string name as the variable
I am writing a custom function that uses an R-function from the
reshape package: cast. However, my question could be applicable to
any R function.
Normally one writes the arguments directly into a function, e.g.:
result=cast(table1, column1 + column2 + column3 ~ column4,
mean) (1)
I need to be able to write this statement as follows:
result=cast(table1, string_with_columns ~
2024 Dec 12
1
Cores hang when calling mcapply
Hi Gregg.
Just wanted to follow up on the solution you proposed.
I had to make some adjustments to get exactly what I wanted, but it works, and takes about 15 minutes on our server configuration:
temp <-
??????open_dataset(
????????????sources = input_files,
????????????format = 'csv',
????????????unify_schema = TRUE,
????????????col_types = schema(
????????????"ID_Key"
2006 Jan 04
3
matrix math
I am using R 2.1.1 in an windows XP environment.
I have 2 dataframes, temp1 and temp2.
Each dataframe has 20 variables (“cocolumns") and 525 observations (“rows”). All variables are numeric.
I want to create a new dataframe that also has 20 columns and 525 rows. The values in this dataframe should be the sum of the 2 other dataframe.
(i.e. temp1$column
2008 Feb 26
1
Split data.frames depeding values of a column
Hello to all
is there a function wich splits a data.frame (column1,column2,column3,....)
into
data1 <-(column1,column3....) #column2 = 1
data2 <-(column1,column3....) #column2 = 2
data3 <-(column1,column3....) #column2 = 3
...
Regards Knut
2024 Dec 12
1
Cores hang when calling mcapply
Hi Thomas,
Glad to hear the suggestion helped, and that switching to a `data.table` approach reduced the processing time and memory overhead?15 minutes for one of the smaller datasets is certainly better! Sounds like the adjustments you devised, especially keeping the multicore approach for `make_clean_names()` and ensuring that `ID_Key` values remain intact, were the missing components you
2024 Dec 11
1
Cores hang when calling mcapply
How is the server configured to handle memory distribution for individual users. I see it has over 700GB of total system memory, but how much can be assigned it each individual user?
AAgain - just curious, and wondering how much memory was assigned to your instance when you were running R.
regards,
Gregg
On Wednesday, December 11th, 2024 at 9:49 AM, Deramus, Thomas Patrick <tderamus at
2006 Jun 07
4
Question: coding protected methods
Apologies first, because I need to ramp up on Ruby and coding Ruby in Rails,
however it''s my 3rd day with this beast :) so I''m asking :
When I added protected methods to the model before it was like:
protected
method....................
end
Would this be a valid way to write a protected method as well ?:
attr_protected :column1, :column2
Perhaps this particular call
2009 Jun 17
6
script help
Hi
?
I have a file. list.txt (two columns)
?
column1??? column2
name??????? address
?
?
I need to put in the letter file letter.txt eg:
?
Dear: Chloe
Address: CA
?
Can I use this
?
for i `cat list.txt` | sed 's/Chloe/$i.1; /CA/$i.2/g' $i.letter.txt
?
Thank you for your help
?
?
?
?
?
__________________________________________________________________
Looking for the perfect gift?
2024 Jun 06
2
R Shiny Help - Trouble passing user input columns to emmeans after ANOVA analysis
Hello everybody,
I have experience coding with R, but am brand new to R Shiny. I am trying
to produce an application that will allow users to upload their own
dataset, select columns they want an ANOVA analysis run on, and generate
graphs that will allow users to view their results. However, I am getting
the following error: *"Argument is of length zero."*
Being new to Shiny, I am
2009 Jun 15
2
Help with syntax error
Hi,
I have written boxplot commands of this form before, but I don''t quite understand why the function call is reporting a syntax error in this instance. All parameters passed to the function are strings.
Thanks in advance.
Payam
> simplevar <- function(wframe,column1,column2) {
+ tframe <- get(wframe)
+ x1 <- which(names(wframe)==column1)
+ x2 <-
2024 Dec 11
1
Cores hang when calling mcapply
Hello Thomas,
Consider that the primary bottleneck may be tied to memory usage and the complexity of pivoting extremely large datasets into wide formats with tens of thousands of unique values per column. Extremely large expansions of columns inherently stress both memory and CPU, and splitting into 110k separate data frames before pivoting and combining them again is likely causing resource
2013 Mar 21
2
Displaying median value over the horizontal(median)line in the boxplot
Hi,
set.seed(45)
test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data
apply(test1,2,function(x) sprintf("%.1f",median(x)))
#columnA columnB
# "44.5"? "10.2"
par(mfrow=c(1,2))
lapply(test1,function(x) {b<-
2004 Oct 25
2
Reading sections of data files based on pattern matching
I am about to write general functions to read the output of simulations
models.
These model generate output files with different sections which I want
to analyze plot etc.
Since this will be used many people at the department I wanted to make
sure that will do this in the best way.
For instance I want to read a snippets of data from a text that look
like this.
2024 Dec 11
1
Cores hang when calling mcapply
About to try this implementation.
As a follow-up, this is the exact error:
Lost warning messages
Error: no more error handlers available (recursive errors?); invoking 'abort' restart
Execution halted
Error: cons memory exhausted (limit reached?)
Error: cons memory exhausted (limit reached?)
Error: cons memory exhausted (limit reached?)
Error: cons memory exhausted (limit reached?)
2010 Feb 27
1
New Variable from Several Existing Variables
I am new to R, but have been using SAS for years. In this transition period,
I am finding myself pulling my hair out to do some of the simplest things.
An example of this is that I need to generate a new variable based on the
outcome of several existing variables in a data row. In other words, if the
variable in all three existing columns are "Yes", then then the new variable
should