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I need to split a vector in a dataframe(last.first) into 2 separate vectors (firstname, lastname) and then put the 2 vectors back into the dataframe. What should I do.
You can split names with strsplit, use whatever is separating the first and second names instead of " " (space in my example).
This will give you a list. Which can be made into dataframe via ldply or unlist to matrix
person.names <- c("Adam Smith", "Max Webber")
temp.list <- strsplit(person.names, " ")
names.df <- ldply(temp.list, function (x) data.frame(first = x[1], second = x[2]))
first second
1 Adam Smith
2 Max Webber
or
matrix(unlist(temp.list), ncol = 2, byrow = TRUE)
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Firstly, I have a big data.frame which has 104 rows and 12 columns, I would like to split it up to 13 rows of 8 rows each with the 12 columns.
I am trying to make a code robust enough to not care how many rows there are but simple make a new data.frame every 8 rows.
Also, is it possible after this point to make a code which loops through the 13 data.frames for some calculations?
Here is a way using data.table.split
library(data.table)
#sample data
set.seed(123)
AA <- data.frame( data = rnorm(104) )
#set number of rows to split on
chunksize = 8
#split on create rowid's
l <- split( setDT(AA)[, rowID := (.I-1) %/% chunksize][], by = "rowID")
#names of the list will become the names of the data.frames
names(l) <- paste0( "df", names(l) )
#write the elements of the list to the global environment, using their names
list2env( l, envir = globalenv() )
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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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If I have a data frame, the standard approach to extracting a value is to use a Boolean/logical expression to match the correct row and column. Example:
set.seed(1)
df <- data.frame(letters = letters[1:3], numbers = as.character(c(1, 2, 1)), value=rnorm(3))
subset(df, letters=="c" & numbers=="1")$value
[1] -0.8356286
However, stringing together many == statements seems a bit kludgey. Another way would be to use row names as keys:
Key <- function(...) paste(..., sep="%") # this could be any formatting
row.names(df) <- with(df, Key(letters, numbers))
df[Key("c", "1"), "value"]
[1] -0.8356286
In the tidyverse, use of row names is discouraged - what would be the recommended way to match and extract values?
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I have a data frame of strings as below and would like to add the string "Market" to each of the elements of the data frame. Is there a function that would allow me to do this easily without having to use a for loop?
V1
1 PUBLIC_DISPATCHSCADA_20141221.zip
2 PUBLIC_DISPATCHSCADA_20141222.zip
3 PUBLIC_DISPATCHSCADA_20141223.zip
4 PUBLIC_DISPATCHSCADA_20141224.zip
5 PUBLIC_DISPATCHSCADA_20141225.zip
6 PUBLIC_DISPATCHSCADA_20141226.zip
We can use paste and specify the delimiter. In this case, I am using _ and pasteing the "Market" at the beginning of the string.
df1$V1 <- paste("Market", df1$V1, sep="_")
If we need to do this for each column
df1[] <- lapply(df1, function(x) paste("Market", x, sep="_"))
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I have my data as below
Idle|Idle|Idle|Idle|Idle|Idle|Idle
Idle|56|55|49|50|53|48|54|52|Idle|Idle|Idle|Idle|Idle|Idle
Idle|49|51|48|50|50|49|50|57|56|57|56|Idle|Idle|69|86|65|Idle|Idle|Idle|Idle
I want to extract numbers in between(which is phone number in ASCII format) which is
(56|55|49|50|53|48|54|52 for 2nd line and 49|51|48|50|50|49|50|57|56|57|56 for 3rd line),
convert them to numbers between "0 and 9" and concatenate as string/number in new column as phone_number in same data set.
2nd row of new column should be 871230652 and 3rd row should be 13022129898
In ASCII format 48 represents 0 and 57 represents 9
Please help
Thanks,
Here's an approach with regular expressions:
res <- sapply(regmatches(x, gregexpr("^(?:Idle\\|)*\\K\\d+(?=\\|)|\\G(?!^)\\|\\K\\d+",
x, perl = TRUE)),
function(x) paste(as.integer(x) - 48, collapse = ""))
# [1] "" "87125064" "13022129898"
If you want to exclude the empty strings, you can use the following command:
res[as.logical(nchar(res))]
# [1] "87125064" "13022129898"
Here x is this vector:
x <- c("Idle|Idle|Idle|Idle|Idle|Idle|Idle",
"Idle|56|55|49|50|53|48|54|52|Idle|Idle|Idle|Idle|Idle|Idle",
"Idle|49|51|48|50|50|49|50|57|56|57|56|Idle|Idle|69|86|65|Idle|Idle|Idle|Idle")