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I want to do an ifelse function here pulling from an existing df "Tokens" a column name called "Vowel" to create column name "Ambiguity".
If column "Vowel" contains "o" or "u", I want to create a column called "High.Ambiguity", and put the value "1"; else, put "0".
What would the syntax for this look like?
I believe this should do the trick for you. mutate creates a new column, in this case called High.Ambiguity which takes on the value 1 when Vowel (a column in Tokens) is either 'o' or 'u' otherwise it is 0.
library(dplyr)
Tokens <- Tokens %>%
mutate(High.Ambiguity = ifelse(Vowel %in% c("o", "u"), 1, 0))
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I have a set of strings in R. In the form of: "X-Y-Z.3000.F.PP0016-C.A-SL-0433.P-N.fC-G.txt". I want to retrieve the set of strings containing just the first occurrence of a string. It depends on the 4th field. In this set for e.g. I have multiple string with X-Y-Z.3000....." I want only the first one having id = 3000, the same for the others.
For reproducibility:
X-Y-Z.3000.F.PP0016-C.A-SL-0433.P-N.fC-G.txt
X-Y-Z.3000.F.PP0016-C.A-SL-0433.F-N.fC-G.txt
X-Y-Z.3008.F.PP0016-C.A-SL-0433.P-N.fC-G.txt
X-Y-Z.3008.F.PP0016-C.B-SX-0433.P-N.fC-G.txt
So at the end I would only the first anche 3th string
X-Y-Z.3000.F.PP0016-C.A-SL-0433.P-N.fC-G.txt
X-Y-Z.3008.F.PP0016-C.A-SL-0433.P-N.fC-G.txt
Extract "4th field" which is 2nd field if we split on ".", then exclude duplicated items:
# data
x <- c("X-Y-Z.3000.F.PP0016-C.A-SL-0433.P-N.fC-G.txt",
"X-Y-Z.3000.F.PP0016-C.A-SL-0433.F-N.fC-G.txt",
"X-Y-Z.3008.F.PP0016-C.A-SL-0433.P-N.fC-G.txt",
"X-Y-Z.3008.F.PP0016-C.B-SX-0433.P-N.fC-G.txt")
x[ !duplicated(sapply(strsplit(x, ".", fixed = "TRUE"), "[", 2)) ]
# [1] "X-Y-Z.3000.F.PP0016-C.A-SL-0433.P-N.fC-G.txt"
# [2] "X-Y-Z.3008.F.PP0016-C.A-SL-0433.P-N.fC-G.txt"
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I have an R list where all of the values are in the first position (i.e. list[1]), while I want all the values to be spread evenly throughout the list (list[1] contains one value, list[2] contains the next, etc.). I have been trying unsuccessfully for a while to split the values one position into separate values (each value is a string of characters separated by spaces) but nothing has worked.
Below is an illustration of the sort of situation I am in.
Say "test" is the name of a list in R. Test is an object of length 1, and if you enter test[1] in the console, the output is thousands of values formatted like so:
[1] "value1" "value2" "value3" ... etc.
Now I want to somehow split the contents of list[1] so that each separated character string is in a separate position, so test[1] is "value1", test[2] is "value2", etc. I have looked around for and attempted many purported solutions to this sort of issue (recent example here: List to integer or double in R) but nothing has worked for me so far.
Here's a simple way:
l1 <- list(l1 = round(rnorm(100, 0, 5), 0))
v <- unlist(l1)
l2 <- as.list(v)
length of l1 is 1 and length of l2 is 100. Is this what you are after?
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How do I extract a number in any given location of a dataframe? Let's say I have a 4x4 matrix, how would I take the number value in (2,4) and assign that value a name?
You can use the setNames function as so: setNames(value, c(name1))
This works for vectors and columns too- for instance: setNames(df[c(col1, col2), c(name1, name2)]; and setNames(c(val1, val2, val3), c(name1, name2, name3))
Edit-
#dataframe with one row and two columns as such
df <- data.frame('a','b')
#You can access a value by:
val <- levels(droplevels(df[1,2])) #Value at first row, second column
#To assign it a name, you can either use:
setNames(val, c(name))
#or
names(val) <- c(name)
Hope this helps!
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Using the first.df data frame, separate the DoB column data into 3 new columns - date, month,year by using the separate() function.I tried last line but it is not giving desired result.
fname <- c("Martina", "Monica", "Stan", "Oscar")
lname <- c("Welch", "Sobers", "Griffith", "Williams")
DoB <- c("1-Oct-1980", "2-Nov-1982", "13-Dec-1979", "27-Jan-1988")
first.df <- data.frame(fname,lname,DoB)
print(first.df)
separate(first.df,DoB,c('date','month','year'),sep = '-')
Moved my comment to an actual answer.
To retain the date column you need to add the remove = FALSE parameter, and to discard one of the separated columns simply add NA instead of a column name. The correct command is then
separate(first.df,DoB,c(NA,'month','year'),sep = '-', remove=FALSE)
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I would like to conditionally replace the values of each columns in R. My data looks like below image.
In this, I want to check if the values are >UCL then want to replace with UCL value and If the values are
The Output should look like below image:
Like this I have many rows and columns data and I'm looking for solution in R.
We create a 'df2' as a copy of 'df1'. Using the 'i1' and 'i2' index, we replace the columns 1:4 in 'df2' with corresponding 'UCL' and 'ICL' values that fits the condition.
df2 <- df1
i1 <- df1[1:4] > df1$UCL
i2 <- df1[1:4] < df1$LCL
df2[1:4][i1] <- df2$UCL[row(df2[1:4])][i1]
df2[1:4][i2] <- df2$LCL[row(df2[1:4])][i2]