R Remove columns from a data.frame with other df - r

Hello I have 2 dataframes:
df1 looks like:
and the df2 looks like:
I have noticied that the df1 has point symbol (.), while df2 has "-". It is weird because both of them, if I open with a text editor or excel, they have "-".
What I need is to drop all the columns of df1 that match with a value of df2.
I have used this:
DataGenSample = df1[,!(names(df1) %in% df2)]
#DataGenSample <- df1[ , !(colnames(df1) %in% df2)]
but there is no change.
All the Data can be foun here. Whith the code that I have used.
# Data (df1):
DataGen <- read.table("data_CNA.txt",sep="\t", header=TRUE, check.names = FALSE)
# Samples (df2):
DeleteSample <- read.table("MuestrasEliminar.txt",sep="\t", header=TRUE, check.names = FALSE)
#Delete columns:
#DataGenSample = DataGen[,!(names(DataGen) %in% DeleteSample)]
DataGenSample <- DataGen[ , !(colnames(DataGen) %in% DeleteSample)]

The issue is - vs ..
When you read in the data, your read command probably has an argument like check.names that changes the names to make them "standard" R names - which means no punctuation other than _ and .. If you set check.names = FALSE the original names will be kept, and your code should work just fine.

Ok, I found out that you need to convert your df in vector first:
vecDeleteSample <- DeleteSample$SAMPLE_ID
And then you can drop the mathing columns of your vector/list:
DataGenSample <- DataGen[,!(names(DataGen) %in% vecDeleteSample)]

Related

Ordering columns of data in R

I have a CSV file with 141 rows and several columns. I wanted my data to be ordered in ascending order by the first two columns i.e. 'label' and 'index'. Following is my code:
final_data <- read.csv("./features.csv",
header = FALSE,
col.names = c('label','index', 'nr_pix', 'rows_with_1', 'cols_with_1',
'rows_with_3p', 'cols_with_3p', 'aspect_ratio',
'neigh_1', 'no_neigh_above', 'no_neigh_below',
'no_neigh_left', 'no_neigh_right', 'no_neigh_horiz',
'no_neigh_vert', 'connected_areas', 'eyes', 'custom'))
sorted_data_by_label <- final_data[order(label),]
sorted_data_by_index <- sorted_data_by_label[order(index),]
write.table(sorted_data_by_index, file = "./features.csv",
append = FALSE, sep = ',',
row.names = FALSE)
I chose to read from a CSV and use write.table because that was necessary for my code requirement to override the CSV with column names.
Now even when I added a , after order(label), and order(index), the code sorted data should still read other rows and columns right?
After running this code, I only get the first row out of 141 rows. Is there a way to fix this problem?
As #akrun has mentioned briefly, what you need to do is to change
sorted_data_by_label <- final_data[order(label),]
to
sorted_data_by_label <- final_data[order(final_data$label),]
and to change
sorted_data_by_index <- sorted_data_by_label[order(index),]
to
sorted_data_by_index <- sorted_data_by_label[order(sorted_data_by_label$index),]
This is because when you write label, R will try to find the index object in the global environment, not within the final_data data frame.
If you intended to use index that is a column of final_data, you need to use explicit final_data$index.
Other options
You can use with:
sorted_data_by_label <- with(final_data, final_data[order(label),])
sorted_data_by_index <- with(sorted_data_by_label, sorted_data_by_label[order(index),])
In dplyr you can use
sorted_data_by_label <- final_data %>% arrange(label)
sorted_data_by_index <- sorted_data_by_label %>% arrange(index)

Filtering process not fetching full data? Using dplyr filter and grep

I have this log file that has about 1200 characters (max) on a line. What I want to do is read this first and then extract certain portions of the file into new columns. I want to extract rows that contain the text “[DF_API: input string]”.
When I read it and then filter based on the rows that I am interested, it almost seems like I am losing data. I tried this using the dplyr filter and using standard grep with the same result.
Not sure why this is the case. Appreciate your help with this. The code and the data is there at the following link.
Satish
Code is given below
library(dplyr)
setwd("C:/Users/satis/Documents/VF/df_issue_dec01")
sec1 <- read.delim(file="secondary1_aa_small.log")
head(sec1)
names(sec1) <- c("V1")
sec1_test <- filter(sec1,str_detect(V1,"DF_API: input string")==TRUE)
head(sec1_test)
sec1_test2 = sec1[grep("DF_API: input string",sec1$V1, perl = TRUE),]
head(sec1_test2)
write.csv(sec1_test, file = "test_out.txt", row.names = F, quote = F)
write.csv(sec1_test2, file = "test2_out.txt", row.names = F, quote = F)
Data (and code) is given at the link below. Sorry, I should have used dput.
https://spaces.hightail.com/space/arJlYkgIev
Try this below code which could give you a dataframe of filtered lines from your file based a matching condition.
#to read your file
sec1 <- readLines("secondary1_aa_small.log")
#framing a dataframe by extracting required lines from above file
new_sec1 <- data.frame(grep("DF_API: input string", sec1, value = T))
names(new_sec1) <- c("V1")
Edit: Simple way to split the above column into multiple columns
#extracting substring in between < & >
new_sec1$V1 <- gsub(".*[<\t]([^>]+)[>].*", "\\1", new_sec1$V1)
#replacing comma(,) with a white space
new_sec1$V1 <- gsub("[,]+", " ", new_sec1$V1)
#splitting into separate columns
new_sec1 <- strsplit(new_sec1$V1, " ")
new_sec1 <- lapply(new_sec1, function(x) x[x != ""] )
new_sec1 <- do.call(rbind, new_sec1)
new_sec1 <- data.frame(new_sec1)
Change columns names for your analysis.

remove rows that a particular column has NA [duplicate]

I am working on a large dataset, with some rows with NAs and others with blanks:
df <- data.frame(ID = c(1:7),
home_pc = c("","CB4 2DT", "NE5 7TH", "BY5 8IB", "DH4 6PB","MP9 7GH","KN4 5GH"),
start_pc = c(NA,"Home", "FC5 7YH","Home", "CB3 5TH", "BV6 5PB",NA),
end_pc = c(NA,"CB5 4FG","Home","","Home","",NA))
How do I remove the NAs and blanks in one go (in the start_pc and end_pc columns)? I have in the past used:
df<- df[-which(is.na(df$start_pc)), ]
... to remove the NAs - is there a similar command to remove the blanks?
df[!(is.na(df$start_pc) | df$start_pc==""), ]
It is the same construct - simply test for empty strings rather than NA:
Try this:
df <- df[-which(df$start_pc == ""), ]
In fact, looking at your code, you don't need the which, but use the negation instead, so you can simplify it to:
df <- df[!(df$start_pc == ""), ]
df <- df[!is.na(df$start_pc), ]
And, of course, you can combine these two statements as follows:
df <- df[!(df$start_pc == "" | is.na(df$start_pc)), ]
And simplify it even further with with:
df <- with(df, df[!(start_pc == "" | is.na(start_pc)), ])
You can also test for non-zero string length using nzchar.
df <- with(df, df[!(nzchar(start_pc) | is.na(start_pc)), ])
Disclaimer: I didn't test any of this code. Please let me know if there are syntax errors anywhere
An elegant solution with dplyr would be:
df %>%
# recode empty strings "" by NAs
na_if("") %>%
# remove NAs
na.omit
Alternative solution can be to remove the rows with blanks in one variable:
df <- subset(df, VAR != "")
An easy approach would be making all the blank cells NA and only keeping complete cases. You might also look for na.omit examples. It is a widely discussed topic.
df[df==""]<-NA
df<-df[complete.cases(df),]

Conditional Insert of Rows

I have a unique dataset, a portion of which can be reproduced using:
data <- textConnection("SNP_Pres,Chr_N,BP_A1F,A1_Beta,A2_SE,ForSortSNP,SortOrder
rs122,13,100461219,C,T,rs122,6
1,16362,0.8701,-0.0048,0.0056,rs122,7
1,19509,0.546015137607046,-0.0033,0.0035,rs122,8
1,17218,0.1539,-0.004,0.013,rs122,9
rs142,13,61952115,G,T,rs142,6
1,16387,0.1295,0.0044,0.0057,rs142,7
1,17218,0.8454,0.006,0.013,rs142,9
rs160,13,100950452,C,T,rs160,6
1,16387,0.549,-0.0021,0.0035,rs160,7
1,19509,0.519102731537216,0.003,0.0027,rs160,8
rs298,13,66664221,C,G,rs298,6
1,19509,0.308290808358246,-0.0032,0.0033,rs298,8
1,17218,0.7227,0.022,0.01,rs298,9")
mydata <- read.csv(data, header = T, sep = ",", stringsAsFactors=FALSE)
It is formatted for use in a program that requires holding spots for missing data entries. In this case, a missing entry is indicated by a numeric skip in the Sort Order column. An entry is complete if the column descends 6 - 7 - 8 - 9, with a new entry beginning again with 6.
I need a way to read through the data file, and insert a row of zeros for each missing entry, so that the file looks like this:
data <- textConnection("SNP_Pres,Chr_N,BP_A1F,A1_Beta,A2_SE,ForSortSNP,SortOrder
rs122,13,100461219,C,T,rs122,6
1,16362,0.8701,-0.0048,0.0056,rs122,7
1,19509,0.546015137607046,-0.0033,0.0035,rs122,8
1,17218,0.1539,-0.004,0.013,rs122,9
rs142,13,61952115,G,T,rs142,6
1,16387,0.1295,0.0044,0.0057,rs142,7
0,0,0,0,0,rs142,8
1,17218,0.8454,0.006,0.013,rs142,9
rs160,13,100950452,C,T,rs160,6
1,16387,0.549,-0.0021,0.0035,rs160,7
1,19509,0.519102731537216,0.003,0.0027,rs160,8
0,0,0,0,0,rs160,9
rs298,13,66664221,C,G,rs298,6
0,0,0,0,0,rs289, 7
1,19509,0.308290808358246,-0.0032,0.0033,rs298,8
1,17218,0.7227,0.022,0.01,rs298,9")
mydata <- read.csv(data, header = T, sep = ",", stringsAsFactors=FALSE)
Ultimately, the last two columns, ForSortSNP and SortOrder will be deleted from the data file, but they are included now for convenience's sake.
Any suggestions are greatly appreicated.
Here is a solution using the expand.grid and merge functions.
grid <- with(mydata, expand.grid(ForSortSNP=unique(ForSortSNP), SortOrder=unique(SortOrder)))
complete <- merge(mydata, grid, all=TRUE, sort=FALSE)
complete[is.na(complete)] <- 0 # replace NAs with 0's
complete <- complete[order(complete$ForSortSNP, complete$SortOrder), ] # re-sort

Rename columns of a data frame by searching column name

I am writing a wrapper to ggplot to produce multiple graphs based on various datasets. As I am passing the column names to the function, I need to rename the column names so that ggplot can understand the reference.
However, I am struggling with renaming of the columns of a data frame
here's a data frame:
df <- data.frame(col1=1:3,col2=3:5,col3=6:8)
here are my column names for search:
col1_search <- "col1"
col2_search <- "col2"
col3_search <- "col3"
and here are column names to replace:
col1_replace <- "new_col1"
col2_replace <- "new_col2"
col3_replace <- "new_col3"
when I search for column names, R sorts the column indexes and disregards the search location.
for example, when I run the following code, I expected the new headers to be new_col1, new_col2, and new_col3, instead the new column names are: new_col3, new_col2, and new_col1
colnames(df)[names(df) %in% c(col3_search,col2_search,col1_search)] <- c(col3_replace,col2_replace,col1_replace)
Does anyone have a solution where I can search for column names and replace them in that order?
require(plyr)
df <- data.frame(col2=1:3,col1=3:5,col3=6:8)
df <- rename(df, c("col1"="new_col1", "col2"="new_col2", "col3"="new_col3"))
df
And you can be creative in making that second argument to rename so that it is not so manual.
> names(df)[grep("^col", names(df))] <-
paste("new", names(df)[grep("^col", names(df))], sep="_")
> names(df)
[1] "new_col1" "new_col2" "new_col3"
If you want to replace an ordered set of column names with an arbitrary character vector, then this should work:
names(df)[sapply(oldNames, grep, names(df) )] <- newNames
The sapply()-ed grep will give you the proper locations for the 'newNames' vector. I suppose you might want to make sure there are a complete set of matches if you were building this into a function.
hmm, this might be way to complicated, but the first that come into my mind:
lookup <- data.frame(search = c(col3_search,col2_search,col1_search),
replace = c(col3_replace,col2_replace,col1_replace))
colnames(df) <- lookup$replace[match(lookup$search, colnames(df))]
I second #justin's aes_string suggestion. But for future renaming you can try.
require(stringr)
df <- data.frame(col1=1:3,col2=3:5,col3=6:8)
oldNames <- c("col1", "col2", "col3")
newNames <- c("new_col1", "new_col2", "new_col3")
names(df) <- str_replace(string=names(df), pattern=oldNames, replacement=newNames)

Resources