I have a huge data frame df, with many columns. One of the columns named id_nm happens to be a character with values such as: aksh123dn.Ins
class(df$id_nm)
returns character
I need to lookup all those values which have the id_nm say aksh123dn.Ins
I used:
new_df<-df[df$id_nm=='aksh123dn.Ins',]
this returns the entire df which isn't the case in reality
also tried:
new_df<-df%filter(id_nm=='aksh123dn.Ins']
still getting the same answer
I think its possibly because it is a character string. Please help me with this. TIA
I have a dataset of airbnb and one of the variables is amenities. The “amenities” column lists all the amenities provided by the host. What’s the total number of amenities offered? Convert this to a numeric value that indicates the number of amenities provided. For example, if an instance of “amenities” is {TV,Internet,Wifi,Washer}, it should convert to 4. Add this as a column to the dataframe. I am very confused on how to do this. Some of the amenities go up to 50 different amenities. Manually making vector would take forever.
I'm also confused on this as well for the airbnb dataset. Before we do any further analysis involving calculations, we should first clean the data for mathematical operations. For example, the character “$” appears in the “price” column, making the data type of “price” character instead of numeric. Remove the “$” and “,” in this column and convert the data type as numeric (modify the raw data). I believe I have to use grep expressions.
if you have that info on a data frame you should try to use strsplit function:
sapply(strsplit(data.frame$amenities,","),length)
for subtitution of characters try gsub function
Say I have multiple data frames, and I want to make a multiple lists of the data frames with the same first column. For example, dfs 1-4 have "abc" in all columns of the first row, dfs 5-7 have "def" in all columns of the first row, etc. How can I write a script which puts (in this case) dfs 1-4 in a list called "abc", dfs 5-7 in a list called "def"?
This is my first question, so please let me know if there is anything else I could provide. I researched for a few days with no luck :(
Thanks!
Jack
So this is a guide to the solution, as you asked.
First make sure you have your list of data frames called l (all(sapply(l, is.data.frame)) should be TRUE).
Then, for each element (df) of this list, you need to get the character (string) in the first row (in any column, for example the first one). This will give you a vector of characters and you can get it by using either sapply or purrr::map_chr.
After that, here comes the split you want to do. Use split for that with as first argument the vectors of indices (see ?seq_along) and as a second argument the vector of characters you've just computed before.
Finally, use lapply to transform this list of indices in a list of data frames (you need to know the [ accessor for a list).
If you need more guidance, don't hesitate to ask.
This is definitely a rookie question but I'm not finding an answer for this (maybe because of my wording) so here goes:
I'm reading a data frame into R studio (csv file) that has 24 columns with headers. There are only numbers in these columns (they're essentially concentrations of several chemicals). It's called all. I need to use them as numeric vectors. When I read them in and type
is.numeric(all[,1])
I get
TRUE
When I type
is.numeric(all[1])
I get
FALSE
I think this is because R interprets the header as a factor. I also tried reading in a table without headers and with headers=FALSE, but R renames it to V1, V2 etc so the result ends up being the same.
I need to work with functions where I invoke something like all[2:24]. How can I go about to make R either "not see" the header or remove it altogether?
Thanks for the answers!
PS: the dataframe I am using (without headers - if it had headers, it would just have names instead of V1, V2, etc) is something like this:
This is a subset from the first column, not the first row.
all[,1]) #subset first column
The following is subset of first row
all[1,]) #subset first row (headers of df not included)
To give columnames
colnames(all) <- c("col1","col2")
Your assumption is wrong. You have a data.frame and all[1] does list subsetting, which results in a data.frame, which is not a vector, and not a numeric vector in particular.
You should study help("[") and An Introduction to R.
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 7 years ago.
Improve this question
I have a huge data set with 20 columns and 20,000 rows, according to the manual of a program I use, we have to put the data as a data frame, though I'm not I understand what it does.. and I can't seem to view the head data frame I created.
I wrote in Bold the part that I don't understand, I'm very new with R, can a kind mind explain to me how the following works?
First I read the CSV file
vData = read.csv("my_matrix.csv");
1) Here we create the data frame as per the manual, what does -c(1:8) do exactly??
dataExpr0 = as.data.frame(t(vData[, -c(1:8)]))
2) Here, to understand what the above part does, I tried to view only the header of the data frame, with the following line, but it display the first 2 columns for the 20,000 rows of data. Is there a way to view only the first 2 rows?
head(dataExpr0, n = 2)
Let's disect what your call is doing, from the inside out.
Basic Indexing
When indexing a data.frame or matrix (assuming 2 dimensions), you access a single element of it with the square bracket notation, as you're seeing. For instance, to see the value in the fourth row, fifth column, you'd use vData[4,5]. This can work with ranges of rows and/or columns as well, such as vData[1:4,5] returning the first 4 rows and the 5th column as a vector.
Note: the range 1:4 can also be an arbitrary vector of numbers, such as vData[c(1,2,5),c(4,8)] which returns a 3 by 2 matrix.
BTW: by default, when the resulting slice/submatrix has one of its dimensions reduced to 1 (as in the latter example), R will drop it to the lower structure (e.g., matrix -> vector -> scalar). In this case, it will drop vData[1:4,5] to a vector. You can prevent this from happening by adding what appears to be a third dimension to the square brackets: vData[1:4,5,drop=FALSE], meaning "do not drop the simplified dimension". Now, you should get a matrix of 4 rows and 1 column in return.
You can read a much more thorough explanation of how to subset data.frames by reading (for example) some of the "Hadleyverse". If you do that, I highly encourage you to make it an interactive session: play in R as you read, to help cement the methods.
Negative Indexing
Negative indices mean "everything except what is listed". In your example, you are subsetting the data to extract everything except columns 1:8. So your vData[,-c(1:8)] is returning all rows and columns 9 through 20, a 20K by 12 matrix. Not small.
Transposition
You probably already know what t() does: transpose the matrix so that it is now 12 by 20K.
A word of warning: if all of your data.frame columns are of the same class (e.g., 'character', 'logical'), then all is fine. However, the fact that data.frames allow disparate types of data in different columns is not a feature shared by matrices. If one data.frame column is different than the others, they will be converted to the highest common format, e.g., logical < integer < numeric < character.
Back to a data.frame
After you transpose it (which converts to a matrix), you convert back to a data.frame, which may or may not be necessary depending on how to intend to deal with the data later. For instance, if the row names are not meaningful, then it may not be that useful to convert into a data.frame. That's relatively immaterial, but I'm a fan of not over-converting things. I'm also a fan of using the simpler data structure, and matrices are typically faster than data.frames.
Head
... merely gives you the top n rows of a data.frame or matrix. In your case, since you transposed it, it is now 20K columns wide, which may be a bit unwieldy on the command line.
Alternatives
Based on what I provided earlier, perhaps you just want to look at the top few rows and first few columns? dataExpr0[1:5,1:5] will work, as will (identically) head(dataExpr0[,1:5], n=5).
More Questions?
I strongly encourage you to read more of the Hadleyverse and become a little more familiar with subsetting and basic data management. It is fundamental to using R, and StackOverflow is not always patient enough to answer baseline questions like this. This forum is best suited for those who have already done some research, read documentation and help pages, and tried some code, and only after that cannot figure out why it is not working. You provided some basic code with is good, but SO is not ideally suited to teach how to start with R.