Given this table
df <- data.frame(col1 = c(letters[3:5], "b","a"),
col2 = c(2:3, 1,1,1))
How can I tell R to return "a".
That means, from the three characters with value of 1 (a tie for the lowest value), I want to select only the first in alphabetical order
I think you want order
with(df, col1[order(col2, col1)][1])
# [1] a
# Levels: a b c d e
or
as.character(with(df, col1[order(col2, col1)][1]))
# [1] "a"
You can order column 1 by the ordered values in column 2 with
df[with(df, order(col2, col1)),]
# col1 col2
# 5 a 1
# 4 b 1
# 3 e 1
# 1 c 2
# 2 d 3
Try:
> min(as.character(df[df$col2==min(df$col2),1]))
[1] "a"
For explanation:
# first find col1 list in rows with minimum of df$col2
> xx = df[df$col2==min(df$col2),1]
> xx
[1] e b a
Levels: a b c d e
# Now find the minimum amongst these after converting factor to character:
> min(as.character(xx))
[1] "a"
>
Related
I have list of lists similar to this sample:
z <- list(list(num1=list((list(tab1=list(list(a=1, b=2, c=5), list(a=3, b=4), list(d=4,e=7)))))),list(num2=list((list(tab2=list(list(a=1, b=2), list(a=3, b=4)))))))
I would like to extract the figures out of the last list of lists names:
Desired output list (since 1 list entries are shorter) or as dataframe with columns corresponding to main list:
[1] a b c a b d e
[2] a b a b
dataframe:
column1 column2
a a
b b
c a
a b
b ""
d ""
e ""
I have tried various combinations of sapply(z, "[[", c("a","b"...) but failed, since the sublist names varies.
EDIT: Sorry, I needed the actual values not the last node (letters)! Additionally, each numeric value has column name, not set in the example above; it is like this:
[[1]]$num1[[1]]$tab1[[1]]$a
Name
1
So the desired solution are values:
[1]
1 2 5 3 4 4 7
[2]
1 2 3 4
I would actually need the numeric values instead of the letters. If you could adjust your solution to this I would be grateful. Thanks.
Try
lapply(z, function(x) as.numeric(unlist(x)))
## [[1]]
## [1] 1 2 5 3 4 4 7
##
## [[2]]
## [1] 1 2 3 4
z1 <- lapply(z, function(x) names(unlist(x)))
z1 <- lapply(z1, function(x) gsub(".*\\.", "", x))
n <- max(sapply(z1, length))
z1 <- lapply(z1, `length<-`, value = n)
setNames(as.data.frame(z1), paste0("Column", seq_along(z1)))
# Column1 Column2
#1 a a
#2 b b
#3 c a
#4 a b
#5 b <NA>
#6 d <NA>
#7 e <NA>
A bit far-fetched and everything but elegant, here is a way to get what you want :
lista<-unlist(lapply(strsplit(names(unlist(z)),"\\."),function(vec) vec[3]))
names(lista)<-unlist(lapply(strsplit(names(unlist(z)),"\\."),function(vec) vec[1]))
uninames<-unique(names(lista))
res<-sapply(uninames,function(x,vec){vec[names(vec)==x]},lista)
> res
$num1
num1 num1 num1 num1 num1 num1 num1
"a" "b" "c" "a" "b" "d" "e"
$num2
num2 num2 num2 num2
"a" "b" "a" "b"
UPDATE
To get the numbers :
a<-unlist(z)
b<-names(unique(z))
res<-sapply(unique(b),function(name,vec,l_name){vec[l_name==name]},a,b)
>res
$num1
num1.tab1.a num1.tab1.b num1.tab1.c num1.tab1.a num1.tab1.b num1.tab1.d num1.tab1.e
1 2 5 3 4 4 7
$num2
num2.tab2.a num2.tab2.b num2.tab2.a num2.tab2.b
1 2 3 4
df1 <- data.frame(
name = c("a", "b", "b", "c"),
score = c(1, 1, 2, 1)
)
How can I get a new data frame with variables/columns from df$name and with each 'corresponding' df$score. I figure that its actually a two-step problem:
First I would need to make a list of (in this example) unequal length vectors like this:
$a
[1] 1
$b
[1] 1 2
$c
[1] 1
Second, NAs need to be padded so one get vectors of equal length before making the desired data frame
that would be like:
a b c
1 1 1 1
2 NA 2 NA
I cannot find any simple means to do this - Im sure there must be!
If the solution can be delivered using dplyr it would be fantastic! Thanks!
To split the data:
(s <- split(df1$score, df1$name))
# $a
# [1] 1
#
# $b
# [1] 1 2
#
# $c
# [1] 1
To create the new data frame:
as.data.frame(sapply(s, `length<-`, max(vapply(s, length, 1L))))
# a b c
# 1 1 1 1
# 2 NA 2 NA
Slightly more efficient would be to use vapply in place of sapply
len <- max(vapply(s, length, 1L))
as.data.frame(vapply(s, `length<-`, double(len), len))
# a b c
# 1 1 1 1
# 2 NA 2 NA
> x <- data.table( C1=c('a','b','c','d') )
> y <- data.table( C1=c('a','b','b','a') )
> f="C1"
> x[ C1 %in% unique(y$C1),]
C1
1: a
2: b
so I can see that the levels of y$C1 cover 2 rows for x$C1.
> y[ C1 %in% unique(x$C1),]
C1
1: a
2: b
3: b
4: a
so I can see that the levels of x$C1 cover 4 rows for y$C1.
This works, but I would like to use a variable for the column name so that I can build a loop when there are many columns.
The following does not work:
> y[ f %in% unique(x$C1),]
Empty data.table (0 rows) of 1 col: C1
This works:
y[ get(f) %in% unique(x$C1),]
the reason for this is that f itself refers to the string "C1"
f
[1] "C1"
class(f)
[1] "character"
you need to refer to the column object "C1" in the data.table itself.
below is an illustration of how get works:
a <- seq(1:10)
b <- "a"
print(b)
[1] "a"
print(get(b))
[1] 1 2 3 4 5 6 7 8 9 10
You could also use:
f <- quote(C1)
y[ eval(f) %in% unique(x$C1),]
# C1
#1: a
#2: b
#3: b
#4: a
I have an example data frame as shown below.
> x=data.frame(id=1:5,c1=letters[1:5],c2=letters[13:17])
> x
id c1 c2
1 1 a m
2 2 b n
3 3 c o
4 4 d p
5 5 e q
I want to create a vector out of this data frame which selects a different column for each row based on another vector. So if that vector is
> vars
[1] 1 2 2 1 1
>
I want for the 1st row in x, column 1, for the second row in x, column 2 and so on. So the expected output vector (or data frame) would be
if vector
a n o d e
if data frame
id V1
1 a
2 n
3 o
4 d
5 e
Any help, much appreciated.
You can 'slice' a data frame using a matrix:
y=data.frame(1:5,c(1,2,2,1,1))
x[2:3][as.matrix(y)]
result:
[1] "a" "n" "o" "d" "e"
Let's generalise this by creating a function
selector=function(x)matrix(c(seq_along(x),x),ncol=2)
Note that there is one column to be ignored at the start, so add 1 to your select vector v
v=c(1,2,2,1,1)
x[selector(v+1)]
result
[1] "a" "n" "o" "d" "e"
Let's say I have a data frame like this:
df <- data.frame(a=letters[1:26],1:26)
And I would like to "re" factor a, b, and c as "a".
How do I do that?
One option is the recode() function in package car:
require(car)
df <- data.frame(a=letters[1:26],1:26)
df2 <- within(df, a <- recode(a, 'c("a","b","c")="a"'))
> head(df2)
a X1.26
1 a 1
2 a 2
3 a 3
4 d 4
5 e 5
6 f 6
Example where a is not so simple and we recode several levels into one.
set.seed(123)
df3 <- data.frame(a = sample(letters[1:5], 100, replace = TRUE),
b = 1:100)
with(df3, head(a))
with(df3, table(a))
the last lines giving:
> with(df3, head(a))
[1] b d c e e a
Levels: a b c d e
> with(df3, table(a))
a
a b c d e
19 20 21 22 18
Now lets combine levels a and e into level Z using recode()
df4 <- within(df3, a <- recode(a, 'c("a","e")="Z"'))
with(df4, head(a))
with(df4, table(a))
which gives:
> with(df4, head(a))
[1] b d c Z Z Z
Levels: b c d Z
> with(df4, table(a))
a
b c d Z
20 21 22 37
Doing this without spelling out the levels to merge:
## Select the levels you want (here 'a' and 'e')
lev.want <- with(df3, levels(a)[c(1,5)])
## now paste together
lev.want <- paste(lev.want, collapse = "','")
## then bolt on the extra bit
codes <- paste("c('", lev.want, "')='Z'", sep = "")
## then use within recode()
df5 <- within(df3, a <- recode(a, codes))
with(df5, table(a))
Which gives us the same as df4 above:
> with(df5, table(a))
a
b c d Z
20 21 22 37
Has anyone tried using this simple method? It requires no special packages, just an understanding of how R treats factors.
Say you want to rename the levels in a factor, get their indices
data <- data.frame(a=letters[1:26],1:26)
lalpha <- levels(data$a)
In this example we imagine we want to know the index for the level 'e' and 'w'
lalpha <- levels(data$a)
ind <- c(which(lalpha == 'e'), which(lalpha == 'w'))
Now we can use this index to replace the levels of the factor 'a'
levels(data$a)[ind] <- 'X'
If you now look at the dataframe factor a there will be an X where there was an e and w
I leave it to you to try the result.
You could do something like:
df$a[df$a %in% c("a","b","c")] <- "a"
UPDATE: More complicated factors.
Data <- data.frame(a=sample(c("Less than $50,000","$50,000-$99,999",
"$100,000-$249,999", "$250,000-$500,000"),20,TRUE),n=1:20)
rows <- Data$a %in% c("$50,000-$99,999", "$100,000-$249,999")
Data$a[rows] <- "$250,000-$500,000"
there are two ways.
if you don't want to drop the unused levels, ie "b" and "c", Joshua's solution is probably best.
if you want to drop the unused levels, then
df$a<-factor(ifelse(df$a%in%c("a","b","c"),"a",as.character(df$a)))
or
levels(df$a)<-ifelse(levels(df$a)%in%c("a","b","c"),"a",levels(df$a))
This is a simplified version of the chosen answer:
I've found that the easiest way to deal with this is to simply overwrite the factor levels by looking at them and then writing the numbers down to be overwritten.
df <- data.frame(a=letters[1:26],1:26)
levels(df)
> [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o"
"p" "q" "r" "s" "t" "u" "v" "w" "x" "y" "z"
levels(df$a)[c(1,2)] <- "c"
summary(df$a)
> c d e f g h i j k l m n o p q r s t u v w x y z
3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1