I have a large table with 50000 obs. The following mimic the structure:
ID <- c(1,2,3,4,5,6,7,8,9)
a <- c("A","B",NA,"D","E",NA,"G","H","I")
b <- c(11,2233,12,2,22,13,23,23,100)
c <- c(12,10,12,23,16,17,7,9,7)
df <- data.frame(ID ,a,b,c)
Where there are some missing values on the vector "a". However, I have some tables where the ID and the missing strings are included:
ID <- c(1,2,3,4,5,6,7,8,9)
a <- c("A","B","C","D","E","F","G","H","I")
key <- data.frame(ID,a)
Is there a way to include the missing strings from key into the column a using the ID?
Another options is to use data.tables fast binary join and update by reference capabilities
library(data.table)
setkey(setDT(df), ID)[key, a := i.a]
df
# ID a b c
# 1: 1 A 11 12
# 2: 2 B 2233 10
# 3: 3 C 12 12
# 4: 4 D 2 23
# 5: 5 E 22 16
# 6: 6 F 13 17
# 7: 7 G 23 7
# 8: 8 H 23 9
# 9: 9 I 100 7
If you want to replace only the NAs (not all the joined cases), a bit more complicated implemintation will be
setkey(setDT(key), ID)
setkey(setDT(df), ID)[is.na(a), a := key[.SD, a]]
You can just use match; however, I would recommend that both your datasets are using characters instead of factors to prevent headaches later on.
key$a <- as.character(key$a)
df$a <- as.character(df$a)
df$a[is.na(df$a)] <- key$a[match(df$ID[is.na(df$a)], key$ID)]
df
# ID a b c
# 1 1 A 11 12
# 2 2 B 2233 10
# 3 3 C 12 12
# 4 4 D 2 23
# 5 5 E 22 16
# 6 6 F 13 17
# 7 7 G 23 7
# 8 8 H 23 9
# 9 9 I 100 7
Of course, you could always stick with factors and factor the entire "ID" column and use the labels to replace the values in column "a"....
factor(df$ID, levels = key$ID, labels = key$a)
## [1] A B C D E F G H I
## Levels: A B C D E F G H I
Assign that to df$a and you're done....
Named vectors make nice lookup tables:
lookup <- a
names(lookup) <- as.character(ID)
lookup is now a named vector, you can access each value by lookup[ID] e.g. lookup["2"] (make sure the number is a character, not numeric)
## should give you a vector of a as required.
lookup[as.character(ID_from_big_table)]
Related
Given a data.frame:
foo <- data.frame(ID=1:10, x=1:10)
rownames(foo) <- LETTERS[1:10]
I would like to reorder a subset of rows, defined by their row names. However, I would like to swap the row names of foo as well. I can do
sel <- c("D", "H") # rows to reorder
foo[sel,] <- foo[rev(sel),]
sel.wh <- match(sel, rownames(foo))
rownames(foo)[sel.wh] <- rownames(foo)[rev(sel.wh)]
but that is long and complicated. Is there a simpler way?
We can replace the sel values in rownames with the reverse of sel.
x <- rownames(foo)
foo[replace(x, x %in% sel, rev(sel)), ]
# ID x
#A 1 1
#B 2 2
#C 3 3
#H 8 8
#E 5 5
#F 6 6
#G 7 7
#D 4 4
#I 9 9
#J 10 10
Not as concise as ronak-shah's answer, but you could also use order.
# extract row names
temp <- row.names(foo)
# reset of vector
temp[which(temp %in% sel)] <- temp[rev(which(temp %in% sel))]
# reset order of data.frame
foo[order(temp),]
ID x
A 1 1
B 2 2
C 3 3
H 8 8
E 5 5
F 6 6
G 7 7
D 4 4
I 9 9
J 10 10
As noted in the comments, this relies on the row names following a lexicographical order. In instances where this is not true, we can use match.
# set up
set.seed(1234)
foo <- data.frame(ID=1:10, x=1:10)
row.names(foo) <- sample(LETTERS[1:10])
sel <- c("D", "H")
Now, the rownames are
# initial data.frame
foo
ID x
B 1 1
F 2 2
E 3 3
H 4 4
I 5 5
D 6 6
A 7 7
G 8 8
J 9 9
C 10 10
# grab row names
temp <- row.names(foo)
# reorder vector containing row names
temp[which(temp %in% sel)] <- temp[rev(which(temp %in% sel))]
Using, match along with order
foo[order(match(row.names(foo), temp)),]
ID x
B 1 1
F 2 2
E 3 3
D 6 6
I 5 5
H 4 4
A 7 7
G 8 8
J 9 9
C 10 10
your data frame is small so you can duplicate it then change the value of each raw:
footmp<-data.frame(foo)
foo[4,]<-footemp[8,]
foot{8,]<-footemp[4,]
Bob
This question already has answers here:
Split comma-separated strings in a column into separate rows
(6 answers)
Closed 6 years ago.
I have a data.frame where one of the variables is a vector (or a list), like this:
MyColumn <- c("A, B,C", "D,E", "F","G")
MyDF <- data.frame(group_id=1:4, val=11:14, cat=MyColumn)
# group_id val cat
# 1 1 11 A, B,C
# 2 2 12 D,E
# 3 3 13 F
# 4 4 14 G
I'd like to have a new data frame with as many rows as the vector
FlatColumn <- unlist(strsplit(MyColumn,split=","))
which looks like this:
MyNewDF <- data.frame(group_id=c(rep(1,3),rep(2,2),3,4), val=c(rep(11,3),rep(12,2),13,14), cat=FlatColumn)
# group_id val cat
# 1 1 11 A
# 2 1 11 B
# 3 1 11 C
# 4 2 12 D
# 5 2 12 E
# 6 3 13 F
# 7 4 14 G
In essence, for every factor which is an element of the list of MyColumn (the letters A to G), I want to assign the corresponding values of the list. Every factor appears only once in MyColumn.
Is there a neat way for this kind of reshaping/unlisting/merging? I've come up with a very cumbersome for-loop over the rows of MyDF and the length of the corresponding element of strsplit(MyColumn,split=","). I'm very sure that there has to be a more elegant way.
You can use separate_rows from tidyr:
tidyr::separate_rows(MyDF, cat)
# group_id val cat
# 1 1 11 A
# 2 1 11 B
# 3 1 11 C
# 4 2 12 D
# 5 2 12 E
# 6 3 13 F
# 7 4 14 G
How about
lst <- strsplit(MyColumn, split = ",")
k <- lengths(lst) ## expansion size
FlatColumn <- unlist(lst, use.names = FALSE)
MyNewDF <- data.frame(group_id = rep.int(MyDF$group_id, k),
val = rep.int(MyDF$val, k),
cat = FlatColumn)
# group_id val cat
#1 1 11 A
#2 1 11 B
#3 1 11 C
#4 2 12 D
#5 2 12 E
#6 3 13 F
#7 4 14 G
We can use cSplit from splitstackshape
library(splitstackshape)
cSplit(MyDF, "cat", ",", "long")
# group_id val cat
#1: 1 11 A
#2: 1 11 B
#3: 1 11 C
#4: 2 12 D
#5: 2 12 E
#6: 3 13 F
#7: 4 14 G
We can also use do with base R with strsplit to split the 'cat' column into a list, replicate the sequence of rows of 'MyDF' with the lengths of 'lst', and create the 'cat' column by unlisting the 'lst'.
lst <- strsplit(as.character(MyDF$cat), ",")
transform(MyDF[rep(1:nrow(MyDF), lengths(lst)),-3], cat = unlist(lst))
I've been playing with some data in order to obtain the ratios between two levels within one variable and taking into account two other variables. I've been using the function aggregate(), which is very useful to calculate means and sums. However, I'm stuck when I want to calculate some ratios (divisions).
Here you find a dataframe very similar to my data:
w<-c("A","B","C","D","E","F","A","B","C","D","E","F")
x<-c(1,1,1,1,1,1,2,2,2,2,2,2)
y<-c(3,4,5,6,8,10,3,4,5,7,9,10)
z<-runif(12)
df<-data.frame(w,x,y,z)
df
w x y z
1 A 1 3 0.93767621
2 B 1 4 0.09169992
3 C 1 5 0.49012926
4 D 1 6 0.90886690
5 E 1 8 0.37058120
6 F 1 10 0.83558267
7 A 2 3 0.42670001
8 B 2 4 0.05656252
9 C 2 5 0.70694423
10 D 2 7 0.13634309
11 E 2 9 0.92065671
12 F 2 10 0.56276176
What I want is to obtain the ratios of z from the two levels of x and taking into account the variables w and y. So the level "A" from the variable "w" in the level "3" from the variable "y" should be:
df$z[1]/df$z[7]
With aggregate function should be something like this:
final<-aggregate(z~y:w, data=df)
However, I know that I miss something because in the variable y there are some classes that not appear in the two categories of w (e.g. 7, 8 and 9).
Any help will be welcomed!
We can use data.table. We convert the 'data.frame' to 'data.table' (setDT(df)), grouped by 'w', 'y', if the nrow (.N) is 2, we divide the first value by the second or else return the 'z'. Assign (:=) the output to a new column 'z1'.
library(data.table)
setDT(df)[,z1 :=if(.N==2) z[1]/z[2] else z , by = .(w,y)]
df
# w x y z z1
# 1: A 1 3 0.93767621 2.1975069
# 2: B 1 4 0.09169992 1.6212135
# 3: C 1 5 0.49012926 0.6933068
# 4: D 1 6 0.90886690 0.9088669
# 5: E 1 8 0.37058120 0.3705812
# 6: F 1 10 0.83558267 1.4847894
# 7: A 2 3 0.42670001 2.1975069
# 8: B 2 4 0.05656252 1.6212135
# 9: C 2 5 0.70694423 0.6933068
#10: D 2 7 0.13634309 0.1363431
#11: E 2 9 0.92065671 0.9206567
#12: F 2 10 0.56276176 1.4847894
If we just want the summary output we don't need to use :=
setDT(df)[, list(z=if(.N==2) z[1]/z[2] else z) , by = .(w,y)]
Or using aggregate
aggregate(z~w+y, df, FUN=function(x)
if(length(x)==2) x[1]/x[2] else x)
This question already has an answer here:
R programming - data frame manoevur
(1 answer)
Closed 7 years ago.
Suppose I have the following dataframe:
Categories Variable
1 a 11
2 b 21
3 c 34
4 d 45
5 e 52
6 f 65
7 g 76
8 a 13
9 b 24
I'd like to turn it into a new dataframe like the following:
Categories Variable
1 a 11
2 b 21
3 c 34
4 d+e 97
5 f 65
6 g 76
7 a 13
8 b 24
How can I do it? (Surely, the dataframe is much larger, but I want the sum of all categories of d and e and group it into a new category, say 'H').
Many thanks!
This is a good question but unfortunately OT here. So I'll answer until it get migrated.
I'm assuming Variable is of class factor, so you'll need to properly re-level it (assuming your data is called df)
levels(df$Categories)[levels(df$Categories) %in% c("d", "e")] <- "h"
Next, I'll use the data.table package as you have a large data set and it's devel version (v >= 1.9.5) has a convinient function called rleid (download from GitHub)
library(data.table) ## v >= 1.9.5
setDT(df)[, .(Variable = sum(Variable)), by = .(indx = rleid(Categories), Categories)]
# indx Categories Variable
# 1: 1 a 11
# 2: 2 b 21
# 3: 3 c 34
# 4: 4 h 97
# 5: 5 f 65
# 6: 6 g 76
# 7: 7 a 13
# 8: 8 b 24
You can try this:
# plyr package provides rbind.fill() function for row binding
library(plyr)
# Assuming you have a rows.cvs containing the data, read it into a data frame
data<-read.csv("rows.csv",stringsAsFactors=FALSE)
# Find the lowest index of d or e (whichever comes first)
index<-min(match("d",data$Var1.nominal.), match("e",data$Var1.nominal.))
# Returns all rows containing d and e in Var1(nominal) column
tempData<-data[data$Var1.nominal. %in% c("d","e"),]
# Remove all the rows containing d and e from original data frame
data<-data[!data$Var1.nominal. %in% c("d","e"),]
# Reorder row index numbers in data
rownames(data)<-NULL
# Combine rows containing d and e in Var1(nominal)column, and sum up the column Var2(numeric)
tempData<-data.frame(Var1.nominal.="d+e",Var2.numeric.=sum(tempData[,2]))
# Combine original data and tempData frame with use of index
data<-rbind.fill(data[1:(index-1),],tempData,data[index:length(data[,1]),])
# Renaming "d+e" to"h"
data[index,1]="h"
# Getting rid of the tempData data frame
rm(tempData)
Output:
> data
Var1.nominal. Var2.numeric.
1 a 11
2 b 21
3 c 34
4 h 97
5 f 65
6 g 76
7 a 13
8 b 24
Hoping there's a simple answer here but I can't find it anywhere.
I have a numeric matrix with row names and column names:
# 1 2 3 4
# a 6 7 8 9
# b 8 7 5 7
# c 8 5 4 1
# d 1 6 3 2
I want to melt the matrix to a long format, with the values in one column and matrix row and column names in one column each. The result could be a data.table or data.frame like this:
# col row value
# 1 a 6
# 1 b 8
# 1 c 8
# 1 d 1
# 2 a 7
# 2 c 5
# 2 d 6
...
Any tips appreciated.
Use melt from reshape2:
library(reshape2)
#Fake data
x <- matrix(1:12, ncol = 3)
colnames(x) <- letters[1:3]
rownames(x) <- 1:4
x.m <- melt(x)
x.m
Var1 Var2 value
1 1 a 1
2 2 a 2
3 3 a 3
4 4 a 4
...
The as.table and as.data.frame functions together will do this:
> m <- matrix( sample(1:12), nrow=4 )
> dimnames(m) <- list( One=letters[1:4], Two=LETTERS[1:3] )
> as.data.frame( as.table(m) )
One Two Freq
1 a A 7
2 b A 2
3 c A 1
4 d A 5
5 a B 9
6 b B 6
7 c B 8
8 d B 10
9 a C 11
10 b C 12
11 c C 3
12 d C 4
Assuming 'm' is your matrix...
data.frame(col = rep(colnames(m), each = nrow(m)),
row = rep(rownames(m), ncol(m)),
value = as.vector(m))
This executes extremely fast on a large matrix and also shows you a bit about how a matrix is made, how to access things in it, and how to construct your own vectors.
A modification that doesn't require you to know anything about the storage structure, and that easily extends to high dimensional arrays if you use the dimnames, and slice.index functions:
data.frame(row=rownames(m)[as.vector(row(m))],
col=colnames(m)[as.vector(col(m))],
value=as.vector(m))