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.
Related
first question, I'll try to go straight to the point.
I'm currently working with tables and I've chosen R because it has no limit with dataframe sizes and can perform several operations over the data within the tables. I am happy with that, as I can manipulate it at my will, merges, concats and row and column manipulation works fine; but I recently had to run a loop with 0.00001 sec/instruction over a 6 Mill table row and it took over an hour.
Maybe the approach of R was wrong to begin with, and I've tried to look for the most efficient ways to run some operations (using list assignments instead of c(list,new_element)) but, since as far as I can tell, this is not something that you can optimize with some sort of algorithm like graphs or heaps (is just tables, you have to iterate through it all) I was wondering if there might be some other instructions or other basic ways to work with tables that I don't know (assign, extract...) that take less time, or configuration over RStudio to improve performance.
This is the loop, just so if it helps to understand the question:
my_list <- vector("list",nrow(table[,"Date_of_count"]))
for(i in 1:nrow(table[,"Date_of_count"])){
my_list[[i]] <- format(as.POSIXct(strptime(table[i,"Date_of_count"]%>%pull(1),"%Y-%m-%d")),format = "%Y-%m-%d")
}
The table, as aforementioned, has over 6 Mill rows and 25 variables. I want the list to be filled to append it to the table as a column once finished.
Please let me know if it lacks specificity or concretion, or if it just does not belong here.
In order to improve performance (and properly work with R and tables), the answer was a mixture of the first comments:
use vectors
avoid repeated conversions
if possible, avoid loops and apply functions directly over list/vector
I just converted the table (which, realized, had some tibbles inside) into a dataframe and followed the aforementioned keys.
df <- as.data.frame(table)
In this case, by doing this the dates were converted directly to character so I did not have to apply any more conversions.
New execution time over 6 Mill rows: 25.25 sec.
I have a small problem, which I don't think is too hard, but I couldn't find any answer here (maybe I phrased my research wrong so please excuse me if the question has already been asked!)
I am importing data from an excel sheet which is split in two columns as in the following picture:
Now, I am trying to import all the data in the second column to my R script, but by splitting it into different vectors: one vector for category A, one for category B, etc... by keeping the data points in the order they are in the file (because as it happens, they are in chronological order).
Now, the categories each have a different number of elements, however, they are ordered alphabetically (ie you'll never find an A in the B's, for example). So I guess that makes it easier, but I'm still a novice with R and I don't really know how to proceed without getting really messy with the code and I know there's probably a simple way of doing it.
Does anyone have an idea on how to treat this nicely please? :)
We can use split in base R to return a list of vectors of 'Data' based on the unique values in 'Category'
lst1 <- split(df1$Data, df1$Category)
Closed. This question is opinion-based. It is not currently accepting answers.
Want to improve this question? Update the question so it can be answered with facts and citations by editing this post.
Closed 5 years ago.
Improve this question
I am learning R through tutorials, but I have difficulties in "how to read" R code, which in turn makes it difficult to write R code. For example:
dir.create(file.path("testdir2","testdir3"), recursive = TRUE)
vs
dup.names <- as.character(data.combined[which(duplicated(as.character(data.combined$name))), "name"])
While I know what these lines of code do, I cannot read or interpret the logic of each line of code. Whether I read left to right or right to left. What strategies should I use when reading/writing R code?
dup.names <- as.character(data.combined[which(duplicated(as.character(data.combined$name))), "name"])
Don't let lines of code like this ruin writing R code for you
I'm going to be honest here. The code is bad. And for many reasons.
Not a lot of people can read a line like this and intuitively know what the output is.
The point is you should not write lines of code that you don't understand. This is not Excel, you do not have but 1 single line to fit everything within. You have a whole deliciously large script, an empty canvas. Use that space to break your code into smaller bits that make a beautiful mosaic piece of art! Let's dive in~
Dissecting the code: Data Frames
Reading a line of code is like looking at a face for familiar features. You can read left to right, middle to out, whatever -- as long as you can lock onto something that is familiar.
Okay you see data.combined. You know (hope) it has rows and columns... because it's data!
You spot a $ in the code and you know it has to be a data.frame. This is because only lists and data.frames (which are really just lists) allow you to subset columns using $ followed by the column name. Subset-by the way- just means looking at a portion of the overall. In R, subsetting for data.frames and matrices can be done using single brackets[, within which you will see [row, column]. Thus if we type data.combined[1,2], it would give you the value in row 1 of column 2.
Now, if you knew that the name of column 2 was name you can use data.combined[1,"name"] to get the same output as data.combined$name[1]. Look back at that code:
dup.names <- as.character(data.combined[which(duplicated(as.character(data.combined$name))), "name"])
Okay, so now we see our eyes should be locked on data.combined[SOMETHING IS IN HERE?!]) and slowly be picking out data.combined[ ?ROW? , Oh the "name" column]. Cool.
Finding those ROW values!
which(duplicated(as.character(data.combined$name)))
Anytime you see the which function, it is just giving you locations. An example: For the logical vector a = c(1,2,2,1), which(a == 1) would give you 1 and 4, the location of 1s in a.
Now duplicated is simple too. duplicated(a) (which is just duplicated(c(1,2,2,1))) will give you back FALSE FALSE TRUE TRUE. If we ran which(duplicated(a)) it would return 3 and 4. Now here is a secret you will learn. If you have TRUES and FALSES, you don't need to use the which function! So maybe which was unnessary here. And also as.character... since duplicated works on numbers and strings.
What You Should Be Writing
Who am I to tell you how to write code? But here's my take.
Don't mix up ways of subsetting: use EITHER data.frame[,column] or data.frame$column...
The code could have been written a little bit more legibly as:
dupes <- duplicated(data.combined$name)
dupe.names <- data.combines$name[dupes]
or equally:
dupes <- duplicated(data.combined[,"name"])
dupe.names <- data.combined[dupes,"name"]
I know this was lengthy but I hope it helps.
An easier way to read any code is to break up their components.
dup.names <-
as.character(
data.combined[which(
duplicated(
as.character(
data.combined$name
)
)
), "name"]
)
For each of the functions - those parts with rounded brackets following them e.g. as.character() you can learn more about what they do and how they work by typing ?as.character in the console
Square brackets [] are use to subset data frames, which are stored in your environment (the box to the upper right if you're using R within RStudio contains your values as well as any defined functions). In this case, you can tell that data.combined is the name that has been given to such a data frame in this example (type ?data.frame to find out more about data frames).
"Unwrapping" long lines of code can be daunting at first. Start by breaking it down into parenthesis , brackets, and commas. Parenthesis directly tacked onto a word indicate a function, and any commas that lie within them (unless they are part of another nested function or bracket) separate arguments which contain parameters that modify the way the function behaves. We can reduce your 2nd line to an outer function as.character and its arguments:
dup.names <- as.character(argument_1)
Just from this, we know that dup.names will be assigned a value with the data type "character" off of a single argument.
Two functions in the first line, file.path() and dir.create(), contain a comma to denote two arguments. Arguments can either be a single value or specified with an equal sign. In this case, the output of file.path happens to perform as argument #1 of dir.create().
file.path(argument_1,argument_2)
dir.create(argument_1,argument_2)
Brackets are a way of subsetting data frames, with the general notation of dataframe_object[row,column]. Within your second line is a dataframe object, data.combined. You know it's a dataframe object because of the brackets directly tacked onto it, and knowing this allows you to that any functions internal to this are contributing to subsetting this data frame.
data.combined[row, column]
So from there, we can see that the internal functions within this bracket will produce an output that specifies the rows of data.combined that will contribute to the subset, and that only columns with name "name" will be selected.
Use the help function to start to unpack these lines by discovering what each function does, and what it's arguments are.
I have a tricky question that I'm hoping someone can help me with. I have an output file that looks pretty standard in that there is one value per row, per column - except for one column (excerpt below) that contains multiple entries per row:
4:103806204-103940896,4:103806204-103940896,4:103822084-103940896,4:103806204-103940896
7:27135712-27139877,7:27135712-27139877
2:209030070-209054773
1:16091458-16113084,1:16090993-16101715,1:16085254-16113084
16:70333061-70367735,16:70323669-70367735,16:70333061-70367735,16:70333061-70367735,16:70328735-70367735,16:70328699-70367735,16:70333061-70367735
It would be easy enough to split this column by ',' but then I won't be able to read it into, say, R very easily.
Instead, I'm hoping I can use a simple bit of code to select only the first two values, and then make one column into two, removing the rest. So the above would become the below:
4 103806204
7 27135712
2 209030070
1 16091458
16 70333061
I lose a little bit of info this way, but it makes the data more manageable. Does anyone have any suggestions?
We can use str_extract_all from library(stringr). We extract the numeric elements (\\d+) in a list, convert the 'character' class to numeric and get the first two elements with head, rbind the list elements.
library(stringr)
do.call(rbind, lapply(str_extract_all(df$col, '\\d+'),
function(x) head(as.numeric(x),2)))
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
How to delete a row in R
I can't figure out how to simply remove row (n) from a dataframe in R.
R's documentation and intro manual are so horribly written, they are virtually zero help on this very simple problem.
Also, every explanation i've found here/ on google is for removing rows that contain strings, or duplicates, etc, which have been excessively advanced for my problem and lead me to introduce more bugs and get nowhere. I just want to remove a row.
Thanks in advance for your help.
fyi the list is in the variable eld, which has 5 columns and 33 rows. I would like to remove row 14. I initialized eld with the following command
eld <- read.table("election2012.txt")
so my desired result is
eldNew <- eld(minus row 14)
eldNew <- eld[-14,]
See ?"[" for a start ...
For ‘[’-indexing only: ‘i’, ‘j’, ‘...’ can be logical
vectors, indicating elements/slices to select. Such vectors
are recycled if necessary to match the corresponding extent.
‘i’, ‘j’, ‘...’ can also be negative integers, indicating
elements/slices to leave out of the selection.
(emphasis added)
edit: looking around I notice
How to delete the first row of a dataframe in R? , which has the answer ... seems like the title should have popped to your attention if you were looking for answers on SO?
edit 2: I also found How do I delete rows in a data frame? , searching SO for delete row data frame ...
Also http://rwiki.sciviews.org/doku.php?id=tips:data-frames:remove_rows_data_frame