问题
I have a large data.frame of character data that I want to convert based on what is commonly called a dictionary in other languages.
Currently I am going about it like so:
foo <- data.frame(snp1 = c(\"AA\", \"AG\", \"AA\", \"AA\"), snp2 = c(\"AA\", \"AT\", \"AG\", \"AA\"), snp3 = c(NA, \"GG\", \"GG\", \"GC\"), stringsAsFactors=FALSE)
foo <- replace(foo, foo == \"AA\", \"0101\")
foo <- replace(foo, foo == \"AC\", \"0102\")
foo <- replace(foo, foo == \"AG\", \"0103\")
This works fine, but it is obviously not pretty and seems silly to repeat the replace statement each time I want to replace one item in the data.frame.
Is there a better way to do this since I have a dictionary of approximately 25 key/value pairs?
回答1:
map = setNames(c("0101", "0102", "0103"), c("AA", "AC", "AG"))
foo[] <- map[unlist(foo)]
assuming that map covers all the cases in foo. This would feel less like a 'hack' and be more efficient in both space and time if foo were a matrix (of character()), then
matrix(map[foo], nrow=nrow(foo), dimnames=dimnames(foo))
Both matrix and data frame variants run afoul of R's 2^31-1 limit on vector size when there are millions of SNPs and thousands of samples.
回答2:
If you're open to using packages, plyr is a very popular one and has this handy mapvalues() function that will do just what you're looking for:
foo <- mapvalues(foo, from=c("AA", "AC", "AG"), to=c("0101", "0102", "0103"))
Note that it works for data types of all kinds, not just strings.
回答3:
Here is a quick solution
dict = list(AA = '0101', AC = '0102', AG = '0103')
foo2 = foo
for (i in 1:3){foo2 <- replace(foo2, foo2 == names(dict[i]), dict[i])}
回答4:
Here's something simple that will do the job:
key <- c('AA','AC','AG')
val <- c('0101','0102','0103')
lapply(1:3,FUN = function(i){foo[foo == key[i]] <<- val[i]})
foo
snp1 snp2 snp3
1 0101 0101 <NA>
2 0103 AT GG
3 0101 0103 GG
4 0101 0101 GC
lapply will output a list in this case that we don't actually care about. You could assign the result to something if you like and then just discard it. I'm iterating over the indices here, but you could just as easily place the key/vals in a list themselves and iterate over them directly. Note the use of global assignment with <<-.
I tinkered with a way to do this with mapply but my first attempt didn't work, so I switched. I suspect a solution with mapply is possible, though.
回答5:
Note this answer started as an attempt to solve the much simpler problem posted in How to replace all values in data frame with a vector of values?. Unfortunately, this question was closed as duplicate of the actual question. So, I'll try to suggest a solution based on replacing factor levels for both cases, here.
In case there is only a vector (or one data frame column) whose values need to be replaced and there are no objections to use factor we can coerce the vector to factor and change the factor levels as required:
x <- c(1, 1, 4, 4, 5, 5, 1, 1, 2)
x <- factor(x)
x
#[1] 1 1 4 4 5 5 1 1 2
#Levels: 1 2 4 5
replacement_vec <- c("A", "T", "C", "G")
levels(x) <- replacement_vec
x
#[1] A A C C G G A A T
#Levels: A T C G
Using the forcatspackage this can be done in a one-liner:
x <- c(1, 1, 4, 4, 5, 5, 1, 1, 2)
forcats::lvls_revalue(factor(x), replacement_vec)
#[1] A A C C G G A A T
#Levels: A T C G
In case all values of multiple columns of a data frame need to be replaced, the approach can be extended.
foo <- data.frame(snp1 = c("AA", "AG", "AA", "AA"),
snp2 = c("AA", "AT", "AG", "AA"),
snp3 = c(NA, "GG", "GG", "GC"),
stringsAsFactors=FALSE)
level_vec <- c("AA", "AC", "AG", "AT", "GC", "GG")
replacement_vec <- c("0101", "0102", "0103", "0104", "0302", "0303")
foo[] <- lapply(foo, function(x) forcats::lvls_revalue(factor(x, levels = level_vec),
replacement_vec))
foo
# snp1 snp2 snp3
#1 0101 0101 <NA>
#2 0103 0104 0303
#3 0101 0103 0303
#4 0101 0101 0302
Note that level_vec and replacement_vec must have equal lengths.
More importantly, level_vec should be complete , i.e., include all possible values in the affected columns of the original data frame. (Use unique(sort(unlist(foo))) to verify). Otherwise, any missing values will be coerced to <NA>. Note that this is also a requirement for Martin Morgans's answer.
So, if there are only a few different values to be replaced you will be probably better off with one of the other answers, e.g., Ramnath's.
回答6:
We can also use dplyr::case_when
library(dplyr)
foo %>%
mutate_all(~case_when(. == "AA" ~ "0101",
. == "AC" ~ "0102",
. == "AG" ~ "0103",
TRUE ~ .))
# snp1 snp2 snp3
#1 0101 0101 <NA>
#2 0103 AT GG
#3 0101 0103 GG
#4 0101 0101 GC
It checks the condition and replaces with the corresponding value if the condition is TRUE. We can add more conditions if needed and with TRUE ~ . we keep the values as it is if none of the condition is matched. If we want to change them to NA instead we can remove the last line.
foo %>%
mutate_all(~case_when(. == "AA" ~ "0101",
. == "AC" ~ "0102",
. == "AG" ~ "0103"))
# snp1 snp2 snp3
#1 0101 0101 <NA>
#2 0103 <NA> <NA>
#3 0101 0103 <NA>
#4 0101 0101 <NA>
This will change the values to NA if none of the above condition is satisfied.
Another option using only base R is to create a lookup dataframe with old and new values, unlist the dataframe, match them with old values, get the corresponding new values and replace.
lookup <- data.frame(old_val = c("AA", "AC", "AG"),
new_val = c("0101", "0102", "0103"))
foo[] <- lookup$new_val[match(unlist(foo), lookup$old_val)]
回答7:
Used @Ramnath's answer above, but made it read (what to be replaced and what to be replaced with) from a file and use gsub rather than replace.
hrw <- read.csv("hgWords.txt", header=T, stringsAsFactor=FALSE, encoding="UTF-8", sep="\t")
for (i in nrow(hrw))
{
document <- gsub(hrw$from[i], hrw$to[i], document, ignore.case=TRUE)
}
hgword.txt contains the following tab separated
"from" "to"
"AA" "0101"
"AC" "0102"
"AG" "0103"
回答8:
Since it's been a few years since the last answer, and a new question came up tonight on this topic and a moderator closed it, I'll add it here. The poster has a large data frame containing 0, 1, and 2, and wants to change them to AA, AB, and BB respectively.
Use plyr:
> df <- data.frame(matrix(sample(c(NA, c("0","1","2")), 100, replace = TRUE), 10))
> df
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1 2 <NA> 2 1 2 0 2 0 2
2 0 2 1 1 2 1 1 0 0 1
3 1 0 2 2 1 0 <NA> 0 1 <NA>
4 1 2 <NA> 2 2 2 1 1 0 1
... to 10th row
> df[] <- lapply(df, as.character)
Create a function over the data frame using revalue to replace multiple terms:
> library(plyr)
> apply(df, 2, function(x) {x <- revalue(x, c("0"="AA","1"="AB","2"="BB")); x})
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
[1,] "AB" "BB" NA "BB" "AB" "BB" "AA" "BB" "AA" "BB"
[2,] "AA" "BB" "AB" "AB" "BB" "AB" "AB" "AA" "AA" "AB"
[3,] "AB" "AA" "BB" "BB" "AB" "AA" NA "AA" "AB" NA
[4,] "AB" "BB" NA "BB" "BB" "BB" "AB" "AB" "AA" "AB"
... and so on
回答9:
Using dplyr::recode:
library(dplyr)
mutate_all(foo, funs(recode(., "AA" = "0101", "AC" = "0102", "AG" = "0103",
.default = NA_character_)))
# snp1 snp2 snp3
# 1 0101 0101 <NA>
# 2 0103 <NA> <NA>
# 3 0101 0103 <NA>
# 4 0101 0101 <NA>
来源：https://stackoverflow.com/questions/7547597/dictionary-style-replace-multiple-items