R How to Remove Rows in a Data Frame Based on the First Character of a Column

R how to remove rows in a data frame based on the first character of a column

We can use grep. The regex ^ indicates the beginning of the string. We match numeric element ([0-9]) at the beginning of the string in the 'y' column using grep. The output will be numeric index, which we use to subset the rows of the 'abc'.

 abc[grep('^[0-9]', abc$y),]
 #        y          z
 #1 34TA912 23.12.2015
 #4 34CC515 25.12.2015

How can i remove rows by condition (initial letters) in r?

You can also convert the data to a data frame and simply use filter() by the dpylr package. In addition to your code, you can have a solution like this:

Data <- data.frame(Data)

Then assign the filtered data to another dataframe.

DataFiltered <- Data %>% filter(Abbreviation %like% "SR_")

Similarly, you can use str_detect() function from stringr package for filtering. This works better, if there is a possibility that Abbreviation has entities having 'SR_' not only as the first 3 characters. You can use RegEx to specify that each entity in filtered data must have Abbreviation starting with 'SR_'.

DataFiltered <- Data %>% filter(str_detect(Abbreviation, pattern = "^SR_"))

Remove Rows occurring after a String R Data frame

We can subset with row_numberand which

library(dplyr)

df %>% filter(row_number() < which(A=='total'))

             A     B
1    Bob Smith 01005
2   Carl Jones 01008
3 Syndey Lewis 01185

Remove first X characters in a data.frame column

1. Use df$c1 <- gsub("Size=", "", df$c1) on both columns to remove the characters
2. Use df$c1 <- as.numeric(as.character(df$c1)) on both columns to change to numeric
3. df$c3 <- df$c1*df$c2 will work to multiply the columns and create a new column of answers

How to remove the first three characters from every row in a column in R

You can do it with gsub function and simple regex. Here is the code:

# Fake data frame
df <- data.frame(text_col = c("abcd", "abcde", "abcdef"))
df$text_col <- as.character(df$text_col)

# Replace first 3 chracters with empty string ""
df$text_col <- gsub("^.{0,3}", "", df$text_col)

R: delete rows that have two columns with the same first digit

You are using str_sub() with the default end parameter, which takes you to the end of the string:. Try this:

x.dist %>% filter(str_sub(county1,1,1)!=str_sub(county2,1,1))

Removing rows from dataframe that contains string in a particular column

There are multiple ways you can do this :

Convert to numeric and remove NA values

subset(df, !is.na(as.numeric(Score)))

#    ID Score
#1 1001     4
#2 1002    20
#5 1005    30

Or with grepl find if there are any non-numeric characters in them and remove them

subset(df, !grepl('\\D', Score))

This can be done with grep as well.

df[grep('\\D', df$Score, invert = TRUE), ]

data

df <- structure(list(ID = 1001:1005, Score = c("4", "20", "h", "v", 
"30")), class = "data.frame", row.names = c(NA, -5L))

Remove Rows From Data Frame where a Row matches a String

Just use the == with the negation symbol (!). If dtfm is the name of your data.frame:

dtfm[!dtfm$C == "Foo", ]

Or, to move the negation in the comparison:

dtfm[dtfm$C != "Foo", ]

Or, even shorter using subset():

subset(dtfm, C!="Foo")

How to remove rows from a data.frame containing a symbol in a particular column

grep!

mydata <- read.table(textConnection("session  first  last  city
9cf571c8faa67cad2aa9ff41f3a26e38  cat biddix  fresno
e30f853d4e54604fd62858badb68113a  caleb+joey  amos  blah
63a5e839510a647c1ff3b8aed684c2a5  me+you  amos  blah"), header=T, stringsAsFactors=FALSE)

grep("\\+",mydata$first)

which returns

[1] 2 3

Telling you in column 2, rows 2 and 3 have a '+' in them.

So you could run:

mydata <- mydata[-grep("\\+",mydata$first),]
mydata

And those entire rows would be deleted. Not sure if it's a typo in your question or not but you say you want to "remove the rows in the first column", do you mean the entries or the entire row?

Delete rows containing specific strings in R

This should do the trick:

df[- grep("REVERSE", df$Name),]

Or a safer version would be:

df[!grepl("REVERSE", df$Name),]

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