I asked this question last week but was looking for how to do it with a batch script. I think it might be possible to do with R, but I'm not very experienced using it. However, after doing some research, I'm pretty sure its impossible to do it with a batch script alone so I think I'll need to use R or VBA script. I don't want someone to just throw the solution at me, if it requires using R or VBA then I'm only interested if you would link some good resources. I've been scouring the interwebs but have found nothing so far.
I need some sort of script that will:
Take the two folders as inputs
Generate a list of all the files in one of them.
For each file:
Read in the data from columns D-G
Find the matching file in the other folder and read in the same data
Compare each cell and verify that the two files match exactly
If they don’t match, report what data doesn’t match
This is what I was asked to do verbatim.
This is what I've done so far.
#echo off
setlocal disableDelayedExpansion
cls
rmdir c:\LocalDirectory/s /q
mkdir c:\LocalDirectory
xcopy "\\SERVER\Path\to\the\files" c:\LocalDirectory
cd c:\LocalDirectory
dir /b /a-d
as you can see, its not much. I can make a list of the files but I need something that can compare them. I know that I could manually do this comparison in Excel but I need to be able to do this for several files. So I'm trying to write some kind of script that will take a specific column of data in a specific excel file in a directory and compare all the rows of that column to a specific column in another xlsx file and then once its done that, generate a message that says either pass or fail. Once thats done, it should move on to the next excel file. The files that have the data thats being compared are named the exact same thing.
Related
Background
I'm doing some data manipulation (joins, etc.) on a very large dataset in R, so I decided to use a local installation of Apache Spark and sparklyr to be able to use my dplyr code to manipulate it all. (I'm running Windows 10 Pro; R is 64-bit.) I've done the work needed, and now want to output the sparklyr table to a .csv file.
The Problem
Here's the code I'm using to output a .csv file to a folder on my hard drive:
spark_write_csv(d1, "C:/d1.csv")
When I navigate to the directory in question, though, I don't see a single csv file d1.csv. Instead I see a newly created folder called d1, and when I click inside it I see ~10 .csv files all beginning with "part". Here's a screenshot:
The folder also contains the same number of .csv.crc files, which I see from Googling are "used to store CRC code for a split file archive".
What's going on here? Is there a way to put these files back together, or to get spark_write_csv to output a single file like write.csv?
Edit
A user below suggested that this post may answer the question, and it nearly does, but it seems like the asker is looking for Scala code that does what I want, while I'm looking for R code that does what I want.
I had the exact same issue.
In simple terms, the partitions are done for computational efficiency. If you have partitions, multiple workers/executors can write the table on each partition. In contrast, if you only have one partition, the csv file can only be written by a single worker/executor, making the task much slower. The same principle applies not only for writing tables but also for parallel computations.
For more details on partitioning, you can check this link.
Suppose I want to save table as a single file with the path path/to/table.csv. I would do this as follows
table %>% sdf_repartition(partitions=1)
spark_write_csv(table, path/to/table.csv,...)
You can check full details of sdf_repartition in the official documentation.
Data will be divided into multiple partitions. When you save the dataframe to CSV, you will get file from each partition. Before calling spark_write_csv method you need to bring all the data to single partition to get single file.
You can use a method called as coalese to achieve this.
coalesce(df, 1)
I'm not sure how to properly ask this but basically I have a very populated single 400 line file on a kaggle competition I was working on and I want to split it up into multiple files (say one file is for data cleaning, another file is for feature engineering etc) in such a way that I can have one main file that will go from reading the csv files all the way to making the model predictions, how can I do that in R? Do I have to encapsulate the entire files into one function each and then use that? If so how does that work? Thanks in advance
You can use the source command and pass it the filename. try ?source
I have two R scripts. The first reads csv files, cleans the data, checks for mathematical errors and corrects them ("errorcheck.R"). The second script gets the clean data from the first, combines column names, expressions and values and creates csv files ("createTables.R").
Originally, the first script was created for importing 5 csv files. But for some projects I might have only 4 or 3 csv files to import, which is fine for the final output. But that throws me an error and when I try to source the first script from the second script, I don't get the clean csv files. How can I source the clean datasets from the first script, even with errors? The errors come only from calling csv files that don't exist.
I'm not sure if this is the same question as:
Is there a way to `source()` and continue after an error?
Can I have some ideas on this please?
Thanks in advance
I am not sure if this serves your answer or not:
Situation:
1. According to your description, your first scripts is made for static input of length of 5. (i.e. 5 .csv file input)
Solution:
I don't know how you take the input of .csv files in first script. I suggest to create a vector of string and pass that to first script and calculate the length of vector to decide how many times your operation should run. Now, the input can be of any length.
So, You can effectively handle any range of .csv files rather than only for 5. Try avoiding hard-coding.
Please let me know if this answer your question. If you face any diffculty just let me know.
Situation
I wrote an R program which I split up into multiple R-files for the sake of keeping a good code structure.
There is a Main.R file which references all the other R-files with the 'source()' command, like this:
source(paste(getwd(), dirname1, 'otherfile1.R', sep="/"))
source(paste(getwd(), dirname3, 'otherfile2.R', sep="/"))
...
As you can see, the working directory needs to be set correctly in advance, otherwise, this could go wrong.
Now, if I want to share this R program with someone else, I have to pass all the R files and folders in relative order of each other for things to work. Hence my next question.
Question
Is there a way to replace all the 'source' commands with the actual R script code which it refers to? That way, I have a SINGLE R script file, which I can simply pass along without having to worry about setting the working directory.
I'm not looking for a solution which is an 'R package' (which by the way is one single directory, so I would lose my own directory structure). I simply wondering if there is an easy way to combine these self-referencing R files into one single file.
Thanks,
Ok I think you could use something like scaning all the files and then writting them again in the same new one. This can be done using readLines and sink:
sink("mynewRfile.R")
for(i in Nfiles){
current_file = readLines(filedir[i])
cat("\n\n#### Current file:",filedir[i],"\n\n")
cat(current_file, sep ="\n")
}
sink()
Here I have supposed all your file directories are in a vector filedir with length Nfiles, I guess you can adapt that
I am EXTREMELY new to R, and programming in general, so thank you for your patience.
I am trying to write a script which reads values from a .txt file and after some manipulation plots the results. I have two questions which are somewhat coupled.
First, is there a function which asks the user to identify the location of a file? i.e. User runs script. Script opens up file navigation prompt and requests user to navigate to and select relevant file.
Currently, I have to manually identify the file and location in R. e.g.
spectra.raw <- read.table("C:\Users\...\file1.txt", row.names=NULL, header = TRUE)
I'd rather have the user identify the file location each time the script is run. This will be used by non-tech people, and I don't trust them to copy/paste file locations into R.
The second question I've been struggling with is, is it possible to create a variable name based off the file selected? For example, if the user selects "file1.txt" I'd like R to assign the output of read.table() to a variable named "file1.raw" much like the above "spectra.raw"
If it helps, all the file names will have the exact same number of characters, so if it's possible to select the last say 5 characters from the file location, that would work.
Thank you very much, and please excuse my ignorance.
See file.choose. Though I believe it behaves slightly differently on different platforms, so beware of that.
See assign, i.e. assign("fileName",value). You'll want to parse the file path that file.choose spits back using string manipulation functions like substr or strsplit.
Try
file.choose
I think it can do what you want.
For example,
myfile <- file.choose()
Enter file name: adataset.Rdata
load(myfile)
myfile contains the name of the file so you don't have to do anything special.