I created a data set using a random row generator:
training_data <- fulldata[sample(nrow(fulldata),100,]
I am under the impression that I can create a second data set of the rest of the data ... rest_data <- fulldata[-training_data] is the code I jotted down in my notes but I am getting
"Error in '[.default'(fulldata, -training_data) :
What part of my code is incorrect?
assuming that fulldatais a dataframe you need a comma in the subscript to indicate that you want the rows of the data frame (i.e. fulldata[rows,columns]). But the indices of the new dataframe training_data will be numbered 1:100so you need a different sort of indicator that corresponds between training_dataand fulldata to show which rows of fulldata should not be included. What you might do is use the rownames, something like:
rest_data<-fulldata[-which(rownames(fulldata)%in%rownames(training_data)),]
which should tell R to remove the rownames of fulldata that occur in training_data. If you have something like an ID variable that is unique to each row you could also use this
rest_data<-fulldata[-which(fulldata$ID%in%training_data$ID),]
Related
Hello, I have this type of table consisting of a single row and several columns. I have tried a code to extract my KD_PL parameters without success. Do you know a way in R to extract all the KD_PLs and store them in a vector or data frame array?
I tried this:
KDPL <- select("KD_PL.", which(substr(colnames(max_LnData), start=1, stop=6)))
This should do the trick:
library(tidyverse)
KDPL <- max_LnData %>% select(starts_with("KD_PL."))
This function selects all columns from your old dataset starting with "KD_PL." and stores them in a new dataframe KDPL.
If you only want the names of the columns to be saved, you could use the following:
KDPL_names <- colnames(KDPL)
This saves the column names in the vector KDPL_names.
I have two data frames, one containing the predictors and one containing the different categories I want to predict. Both of the data frames contain a column named geoid. Some of the rows of my predictors contains NA values, and I need to remove these.
After extracting the geoid value of the rows containing NA values, and removing them from the predictors data frame I need to remove the corresponding rows from the categories data frame as well.
It seems like a rather basic operation but the code won't work.
categories <- as.data.frame(read.csv("files/cat_df.csv"))
predictors <- as.data.frame(read.csv("files/radius_100.csv"))
NA_rows <- predictors[!complete.cases(predictors),]
geoids <- NA_rows['geoid']
clean_categories <- categories[!(categories$geoid %in% geoids),]
None of the rows in categories/clean_categories are removed.
A typical geoid value is US06140231. typeof(categories$geoid) returns integer.
I can't say this is it, but a very basic typo won't be doing what you want, try this correction
clean_categories <- categories[!(categories$geoid %in% geoids),]
Almost certainly this is what you meant to happen in that line. You want to negate the result of the %in% operator. You don't include a reproducible example so I can't say whether the whole thing will do as you want.
All other questions and answers I have seen on this have not been useful.
I have imported data from an Excel sheet into R and I want to remove all of the non-numeric data from a particular column so I can perform calculations. Ideally I would like to be able to create a function that does this.
Assuming the data frame is called price and the column I want is Q1:
I have found answers to this question where I could do this by using
convert <- as.numeric(as.character(price$Q1)
ncolumn <-concert[!is.na(convert)]
However, when I try and create the function I want to have the inputs to be both the name of the data frame as well as the name of the column. I have tried using price[2] instead of price$Q1 in that first line I showed but it doesn't seem to work. I also tried extracting the name of the column and the name of the data frame and using the $ notation instead of [ ] and it still doesn't work.
Guessing from the question, would the below be what you are trying to do?
#function with inputs of both name of a data frame and name of a column
NumConv <- function(data, column){
convert<-as.numeric(as.character(data[, column]))
convert[!is.na(convert)]
}
#executing the function with your assumed example
NumConv(price, "Q1")
To remove rows from a data frame, I use the following command:
data <- data[-1, ]
for example to remove the first row. I need to remove the first 6 rows, so I used the following:
data <- data[-c(1,2,3,4,5,6), ]
OR
data <- data[-(1:6), ]
this works as far as removing the row names, but introduced a new column called row.names that I cannot get rid of unless I use the command:
row.names(data) <- NULL
What is the reason for this? Is there a better way of removing a number of rows/columns with one command?
Example:
after the following code:
tquery <- tquery[-(1:6), ]
This is the data:
Although it seems as such, you are not actually adding a column to the data. What you are seeing is just a result of using View(). The function is showing the "row.names" attribute of the data frame as the first column, but you didn't really add the column.
This is expected and documented behavior. From the Details section of help(View)
If there are row names on the data frame that are not 1:nrow, they are displayed in a separate first column called row.names.
So since you subsetted the data, the row names are technically not 1:nrow any more and hence the new column is introduced in the viewer.
Print your data in the console and you'll see the difference.
View(mtcars) ## because the mtcars row names are not 1:nrow
versus
mtcars
Basically, don't trust View() to display an exact representation of the actual data. Instead use attributes(), *names(), dim(), length(), etc. or just peek at the data with head().
See r help via "?row.names" for more info. From the documentation, "All data frames have a row names attribute"
?row.names ## get more information about row.names from r help
row.names is not a new column, but rather an attribute of every single data frame. This is simply meta data and is ignored by most data. When you output this data (i.e. CSV) or use it in a function, this data will not interfere. This is similar to how excel has row numbers on the left margin, which is referential data for the application.
str(your_dataframe) ## see that those columns don't exist
colnames(your_dataframe) ## see column names
Is it possible to append a column to data frame in the following scenario?
dfWithData <- data.frame(start=c(1,2,3), end=c(11,22,33))
dfBlank <- data.frame()
..how to append column start from dfWithData to dfBlank?
It looks like the data should be added when data frame is being initialized. I can do this:
dfBlank <- data.frame(dfWithData[1])
but I am more interested if it is possible to append columns to an empty (but inti)
I would suggest simply subsetting the data frame you get back from the RODBC call. Something like:
df[,c('A','B','D')]
perhaps, or you can also subset the columns you want with their numerical position, i.e.
df[,c(1,2,4)]
dfBlank[1:nrow(dfWithData),"start"] <- dfWithData$start