I've got two data frames in which the unique identifiers common to both frames differ in the number of observations. I would like to create a dataframe from both in which the observations from each frame are taken if they have more observations for a common identifier. For example:
f1 <- data.frame(x = c("a", "a", "b", "c", "c", "c"), y = c(1,1,2,3,3,3))
f2 <- data.frame(x = c("a","b", "b", "c", "c"), y = c(4,5,5,6,6))
I would like this to generate a merge based on the longer x such that it produces:
x y
a 1
a 1
b 5
b 5
c 3
c 3
c 3
Any and all thoughts would be great.
Here's a solution using split
dd<-rbind(cbind(f1, s="f1"), cbind(f2, s="f2"))
keep<-unsplit(lapply(split(dd$s, dd$x), FUN=function(x) {
y<-table(x)
x == names(y[which.max(y)])
}), dd$x)
dd <- dd[keep,]
Normally i'd prefer to use the ave function here but because i'm changing data.types from a factor to a logical, it wasn't as appropriate so I basically copied the idea that ave uses and used split.
dplyr solution
library(dplyr)
First we combine the data:
with rbind() and introduce a new variable called ref to know where each observation came from:
both <- rbind( f1, f2 )
both$ref <- rep( c( "f1", "f2" ) , c( nrow(f1), nrow(f2) ) )
then count the observations:
make another new variable that contains how many observations for each ref and x combination:
both_with_counts <- both %>%
group_by( ref ,x ) %>%
mutate( counts = n() )
then filter for the largest count:
both_with_counts %>% group_by( x ) %>% filter( n==max(n) )
note: you could also select only the x and y cols with select(x,y)...
this gives:
## Source: local data frame [7 x 4]
## Groups: x
##
## x y ref counts
## 1 a 1 f1 2
## 2 a 1 f1 2
## 3 c 3 f1 3
## 4 c 3 f1 3
## 5 c 3 f1 3
## 6 b 5 f2 2
## 7 b 5 f2 2
Altogether now...
what_I_want <-
rbind(cbind(f1,ref = "f1"),cbind(f2,ref = "f2")) %>%
group_by(ref,x) %>%
mutate(counts = n()) %>%
group_by( x ) %>%
filter( counts==max(counts) ) %>%
select( x, y )
and thus:
> what_I_want
# Source: local data frame [7 x 2]
# Groups: x
#
# x y
# 1 a 1
# 2 a 1
# 3 c 3
# 4 c 3
# 5 c 3
# 6 b 5
# 7 b 5
Not a elegant answer but still give the desired result. Hope this help.
f1table <- data.frame(table(f1$x))
colnames(f1table) <- c("x","freq")
f1new <- merge(f1,f1table)
f2table <- data.frame(table(f2$x))
colnames(f2table) <- c("x","freq")
f2new <- merge(f2,f2table)
table <- rbind(f1table, f2table)
table <- table[with(table, order(x,-freq)), ]
table <- table[!duplicated(table$x), ]
data <-rbind(f1new, f2new)
merge(data, table, by=c("x","freq"))[,c(1,3)]
x y
1 a 1
2 a 1
3 b 5
4 b 5
5 c 3
6 c 3
7 c 3
Related
My data is:
df <- data.frame(a = 1:2,
x = 1:2,
b = 1:2,
y = 3:4,
x_2 = 1:2,
y_2 = 3:4,
c = 1:2,
x_3 = 5:6,
y_3 = 1:2)
I now want to put together the x vars, and the y vars so that the order of columns would be:
a, x, x_2, x_3, b, y, y_2, y_3, c
I thought, I could use tidyverse's relocate function in combination with lapply or map or reduce (?), but it doesn't work out.
E.g. if I do:
move_names <- c("x", "y")
library(tidyverse)
moved_data <- lapply(as.list(move_names), function(x)
{
df <- df |>
relocate(!!!syms(paste0(x, "_", 2:3)),
.after = all_of(x))
}
)
It does the moving for x and y separately, but it creates separate list, but I want to have just my original df with relocated columns.
Update:
I should have been clear that my real data frame has ~500 columns where the to-be-moved columns are all over the place. So providing the full vector of desired column name order won't be feasible.
What I instead have: I have the names of my original columns, i.e. x and y, and I have the names of the to-be-moved columns, i.e. x_2, x_3, y_2, y_3.
In base R:
df[match(c('a', 'x', 'x_2', 'x_3', 'b', 'y', 'y_2', 'y_3', 'c'), names(df))]
#> a x x_2 x_3 b y y_2 y_3 c
#> 1 1 1 1 5 1 3 3 1 1
#> 2 2 2 2 6 2 4 4 2 2
Not sure if it's what you want.
Vector with order of column names
Let's say you have a vector relocate_name that contains the order of your columns:
library(tidyverse)
relocate_name <- c("a", "x", "x_2", "x_3", "b", "y", "y_2", "y_3", "c")
df %>% relocate(any_of(relocate_name))
Vector with prefix of column names
Or if you only have the prefix of the order, let's call it relocate_name2:
relocate_name2 <- c("a", "x", "b", "y", "c")
df %>% relocate(starts_with(relocate_name2))
Group x and y together
Or if you only want to "group" x and y together:
df %>%
relocate(starts_with("x"), .after = "x") %>%
relocate(starts_with("y"), .after = "y")
Output
All of the above output is the same.
a x x_2 x_3 b y y_2 y_3 c
1 1 1 1 5 1 3 3 1 1
2 2 2 2 6 2 4 4 2 2
library(rlist)
# split based in colname-part before _
L <- split.default(df, f = gsub("(.*)_.*", "\\1", names(df)))
# remove names with an underscore
# this is the new order, it should match the names of list L !!
neworder <- names(df)[!grepl("_", names(df))]
# [1] "a" "x" "b" "y" "c"
# cbind list elements together
ans <- rlist::list.cbind(L[neworder])
# a x.x x.x_2 x.x_3 b y.y y.y_2 y.y_3 c
# 1 1 1 1 5 1 3 3 1 1
# 2 2 2 2 6 2 4 4 2 2
# create tidy names again
names(ans) <- gsub(".*\\.(.*)", "\\1", names(ans))
# a x x_2 x_3 b y y_2 y_3 c
# 1 1 1 1 5 1 3 3 1 1
# 2 2 2 2 6 2 4 4 2 2
Ok, this is probably the worst workaround ever and I don't really understand what exactly I'm doing (especially with the <<-), but it is does the trick.
My general idea after realizing the problem a bit more with the help of you guys here was to "loop" through both of my x and y names, remove these new _2 and _3 columns from the vector of column names and re-append them after their "base" x and y columns.
search_names <- c("x", "y")
df_names <- names(df)
new_names <- lapply(search_names, function(x)
{
start <- which(df_names == x)
without_new_names <- setdiff(df_names, paste0(x, "_", 2:3))
df_names <<- append(without_new_names, values = paste0(x, "_", 2:3), after = start)
})[[length(search_names)]]
df |>
relocate(any_of(new_names))
a x x_2 x_3 b y y_2 y_3 c
1 1 1 1 5 1 3 3 1 1
2 2 2 2 6 2 4 4 2 2
There is my problem that I can't solve it:
Data:
df <- data.frame(f1=c("a", "a", "b", "b", "c", "c", "c"),
v1=c(10, 11, 4, 5, 0, 1, 2))
data.frame:f1 is factor
f1 v1
a 10
a 11
b 4
b 5
c 0
c 1
c 2
# What I want is:(for example, fetch data with the number of element of some level == 2, then to data.frame)
a b
10 4
11 5
Thanks in advance!
I might be missing something simple here , but the below approach using dplyr works.
library(dplyr)
nlevels = 2
df1 <- df %>%
add_count(f1) %>%
filter(n == nlevels) %>%
select(-n) %>%
mutate(rn = row_number()) %>%
spread(f1, v1) %>%
select(-rn)
This gives
# a b
# <int> <int>
#1 10 NA
#2 11 NA
#3 NA 4
#4 NA 5
Now, if you want to remove NA's we can do
do.call("cbind.data.frame", lapply(df1, function(x) x[!is.na(x)]))
# a b
#1 10 4
#2 11 5
As we have filtered the dataframe which has only nlevels observations, we would have same number of rows for each column in the final dataframe.
split might be useful here to split df$v1 into parts corresponding to df$f1. Since you are always extracting equal length chunks, it can then simply be combined back to a data.frame:
spl <- split(df$v1, df$f1)
data.frame(spl[lengths(spl)==2])
# a b
#1 10 4
#2 11 5
Or do it all in one call by combining this with Filter:
data.frame(Filter(function(x) length(x)==2, split(df$v1, df$f1)))
# a b
#1 10 4
#2 11 5
Here is a solution using unstack :
unstack(
droplevels(df[ave(df$v1, df$f1, FUN = function(x) length(x) == 2)==1,]),
v1 ~ f1)
# a b
# 1 10 4
# 2 11 5
A variant, similar to #thelatemail's solution :
data.frame(Filter(function(x) length(x) == 2, unstack(df,v1 ~ f1)))
My tidyverse solution would be:
library(tidyverse)
df %>%
group_by(f1) %>%
filter(n() == 2) %>%
mutate(i = row_number()) %>%
spread(f1, v1) %>%
select(-i)
# # A tibble: 2 x 2
# a b
# * <dbl> <dbl>
# 1 10 4
# 2 11 5
or mixing approaches :
as_tibble(keep(unstack(df,v1 ~ f1), ~length(.x) == 2))
Using all base functions (but you should use tidyverse)
# Add count of instances
x$len <- ave(x$v1, x$f1, FUN = length)
# Filter, drop the count
x <- x[x$len==2, c('f1','v1')]
# Hacky pivot
result <- data.frame(
lapply(unique(x$f1), FUN = function(y) x$v1[x$f1==y])
)
colnames(result) <- unique(x$f1)
> result
a b
1 10 4
2 11 5
I'd like code this, may it helps for you
library(reshape2)
library(dplyr)
aa = data.frame(v1=c('a','a','b','b','c','c','c'),f1=c(10,11,4,5,0,1,2))
cc = aa %>% group_by(v1) %>% summarise(id = length((v1)))
dd= merge(aa,cc) #get the level
ee = dd[dd$aa==2,] #select number of level equal to 2
ee$id = rep(c(1,2),nrow(ee)/2) # reset index like (1,2,1,2)
dcast(ee, id~v1,value.var = 'f1')
all done!
I would like to combine a set of data frames into a single data frame by summing columns that have matching variables (instead of appending columns).
For example, given
df1 <- data.frame(A = c(0,0,1,1,1,2,2), B = c(1,2,1,2,3,1,5), x = c(2,3,1,5,3,7,0))
df2 <- data.frame(A = c(0,1,1,2,2,2), B = c(1,1,3,2,4,5), x = c(4,8,4,1,0,3))
df3 <- data.frame(A = c(0,1,2), B = c(5,4,2), x = c(5,3,1))
I want to match by "A" and "B" and sum the values of "x". For this example, I can get the desired result as follows:
library(plyr)
library(dplyr)
# rename columns so that join_all preserves them all:
colnames(df1)[3] <- "x1"
colnames(df2)[3] <- "x2"
colnames(df3)[3] <- "x3"
# join the data frames by matching "A" and "B" values:
res <- join_all(list(df1, df2, df3), by = c("A", "B"), type = "full")
# get the sums and drop superfluous columns:
arrange(res, A, B) %>%
rowwise() %>%
mutate(x = sum(x1, x2, x3, na.rm = TRUE)) %>%
select(A, B, x)
Result:
A B x
<dbl> <dbl> <dbl>
1 0 1 6
2 0 2 3
3 0 5 5
4 1 1 9
5 1 2 5
6 1 3 7
7 1 4 3
8 2 1 7
9 2 2 2
10 2 4 0
11 2 5 3
A more general solution is
library(dplyr)
# function to get the desired result for two data frames:
my_merge <- function(df1, df2)
{
m1 <- merge(df1, df2, by = c("A", "B"), all = TRUE)
m1 <- rowwise(res) %>%
mutate(x = sum(x.x, x.y, na.rm = TRUE)) %>%
select(A, B, x)
return(m1)
}
l1 <- list(df2, df3) # omit the first data frame
res <- df1 # initial value of the result
for(df in l1) res <- my_merge(res, df) # call the function repeatedly
Is there a more efficient option for combining a large set of data frames? Ideally it should be recursive (i.e. it's better not to join all data frames into one massive data frame before calculating the sums).
An easier option is to bind the rows of the datasets, then group by the columns of interest and get the summarised output by getting the sum of 'x'
library(tidyverse)
bind_rows(df1, df2, df3) %>%
group_by(A, B) %>%
summarise(x = sum(x))
# A tibble: 11 x 3
# Groups: A [?]
# A B x
# <dbl> <dbl> <dbl>
# 1 0 1 6
# 2 0 2 3
# 3 0 5 5
# 4 1 1 9
# 5 1 2 5
# 6 1 3 7
# 7 1 4 3
# 8 2 1 7
# 9 2 2 2
#10 2 4 0
#11 2 5 3
If there are many objects in the global environment with the pattern "df" followed by some digits
mget(ls(pattern= "^df\\d+")) %>%
bind_rows %>%
group_by(A, B) %>%
summarise(x = sum(x))
As the OP mentioned about memory constraints, if we do the join first and then use rowSums or + with reduce, it would be more efficient
mget(ls(pattern= "^df\\d+")) %>%
reduce(full_join, by = c("A", "B")) %>%
transmute(A, B, x = rowSums(.[3:5], na.rm = TRUE)) %>%
arrange(A, B)
# A B x
#1 0 1 6
#2 0 2 3
#3 0 5 5
#4 1 1 9
#5 1 2 5
#6 1 3 7
#7 1 4 3
#8 2 1 7
#9 2 2 2
#10 2 4 0
#11 2 5 3
This could also be done with data.table
library(data.table)
rbindlist(mget(ls(pattern= "^df\\d+")))[, .(x = sum(x)), by = .(A, B)]
Ideally it should be recursive (i.e. it's better not to join all data frames into one massive data frame before calculating the sums).
If you're memory constrained and willing to sacrifice speed (vs #akrun's data.table approach), use one table at a time in a loop:
library(data.table)
tabs = c("df1", "df2", "df3")
# enumerate all combos for the results table
# initializing sum to 0
res = CJ(A = 0:2, B = 1:5, x = 0)
# loop over tabs, adding on
for (i in seq_along(tabs)){
tab = get(tabs[[i]])
res[tab, on=.(A, B), x := x + i.x][]
rm(tab)
}
If you need to read tables from disk, change tabs to file names and get to fread or whatever function.
I am skeptical that you can fit all the tables in memory, but cannot also fit an rbind-ed copy of them together.
Similarly (thanks to #akrun's comment), use his approach pairwise:
res = data.table(get(tabs[[1]]))[0L]
for (i in seq_along(tabs)){
tab = get(tabs[[i]])
res = rbind(res, tab)[, .(x = sum(x)), by=.(A,B)]
rm(tab)
}
I want to efficiently sum the entries of two data frames, though the data frames are not guaranteed to have the same dimensions or column names. Merge isn't really what I'm after here. Instead I want to create an output object with all of the row and column names that belong to either of the added data frames. In each position of that output, I want to use the following logic for the computed value:
If a row/column pairing belongs to both input data frames I want the output to include their sum
If a row/column pairing belongs to just one input data frame I want to include that value in the output
If a row/column pairing does not belong to any input matrix I want to have 0 in that position in the output.
As an example, consider the following input data frames:
df1 = data.frame(x = c(1,2,3), y = c(4,5,6))
rownames(df1) = c("a", "b", "c")
df2 = data.frame(x = c(7,8), z = c(9,10), w = c(2, 3))
rownames(df2) = c("a", "d")
> df1
x y
a 1 4
b 2 5
c 3 6
> df2
x z w
a 7 9 2
d 8 10 3
I want the final result to be
> df2
x y z w
a 8 4 9 2
b 2 5 0 0
c 3 6 0 0
d 8 0 10 3
What I've done so far -
bind_rows / bind_cols in dplyr can throw the following:
"Error: incompatible number of rows (3, expecting 2)"
I have duplicated column names, so 'merge' isn't working for my purposes either - returns an empty df for some reason.
Seems like you could merge on the rownames, then take care of the sums and conversion of NA to zero with some additional munging:
library(dplyr)
df.new = df1 %>% add_rownames %>%
full_join(df2 %>% add_rownames, by="rowname") %>%
mutate_each(funs(replace(., which(is.na(.)), 0))) %>%
mutate(x = x.x + x.y) %>%
select(rowname,x,y,z,w)
Or, with #DavidArenburg's much more elegant and extensible solution:
df.new = df1 %>% add_rownames %>%
full_join(df2 %>% add_rownames) %>%
group_by(rowname) %>%
summarise_each(funs(sum(., na.rm = TRUE)))
df.new
rowname x y z w
1 a 8 4 9 2
2 b 2 5 0 0
3 c 3 6 0 0
4 d 8 0 10 3
This seems like some type of a simple merge on common column names (+ row names) and then a simple aggregation, this is how I would tackle this
library(data.table)
merge(setDT(df1, keep.rownames = TRUE), # Convert to data.table + keep rows
setDT(df2, keep.rownames = TRUE), # Convert to data.table + keep rows
by = intersect(names(df1), names(df2)), # merge on common column names
all = TRUE)[, lapply(.SD, sum, na.rm = TRUE), by = rn] # Sum all columns by group
# rn x y z w
# 1: a 8 4 9 2
# 2: b 2 5 0 0
# 3: c 3 6 0 0
# 4: d 8 0 10 3
Are a pretty straight forward base R solution
df1$rn <- row.names(df1)
df2$rn <- row.names(df2)
res <- merge(df1, df2, all = TRUE)
rowsum(res[setdiff(names(res), "rn")], res[, "rn"], na.rm = TRUE)
# x y z w
# a 8 4 9 2
# b 2 5 0 0
# c 3 6 0 0
# d 8 0 10 3
First, I would grab the names of all the rows and columns of the new entity:
(all.rows <- unique(c(row.names(df1), row.names(df2))))
# [1] "a" "b" "c" "d"
(all.cols <- unique(c(names(df1), names(df2))))
# [1] "x" "y" "z" "w"
Then I would construct an output matrix with those rows and column names (with matrix data initialized to all 0s), adding df1 and df2 to the relevant parts of that matrix.
out <- matrix(0, nrow=length(all.rows), ncol=length(all.cols))
rownames(out) <- all.rows
colnames(out) <- all.cols
out[row.names(df1),names(df1)] <- unlist(df1)
out[row.names(df2),names(df2)] <- out[row.names(df2),names(df2)] + unlist(df2)
out
# x y z w
# a 8 4 9 2
# b 2 5 0 0
# c 3 6 0 0
# d 8 0 10 3
Using xtabs on melted / stacked data frames:
out <- rbind(cbind(rn=rownames(df1),stack(df1)), cbind(rn=rownames(df2),stack(df2)))
as.data.frame.matrix(xtabs(values ~ rn + ind, data=out))
# x y w z
#a 8 4 2 9
#b 2 5 0 0
#c 3 6 0 0
#d 8 0 3 10
I’m not convinced the accepted (or alternative merge) method is the best. It will give incorrect results if you have common rows, they’ll get joined and not summed.
This can be shown trivialy by changing df2 to:
df2 = data.frame(x = c(1,2), y = c(4,5), z = c(9,10), w = c(2, 3))
rownames(df2) = c("a", "d")
expected results:
rn x y z w
1: a 2 8 9 2
2: b 2 5 0 0
3: c 3 6 0 0
4: d 2 5 10 3
actual results
merge(setDT(df1, keep.rownames = TRUE),
setDT(df2, keep.rownames = TRUE),
by = intersect(names(df1), names(df2)),
all = TRUE)[, lapply(.SD, sum, na.rm = TRUE), by = rn]
rn x y z w
1: a 1 4 9 2
2: b 2 5 0 0
3: c 3 6 0 0
4: d 2 5 10 3
You need to combine both the outer join with an inner join (or left/right joins, merge all=T/all=F). Or alternatively using plyr’s rbind.fill :
base R solution
res <- rbind.fill(df1,df2)
rowsum(res[setdiff(names(res), "rn")], res[, "rn"], na.rm = TRUE)
data table solution
as.data.table(rbind.fill(
setDT(df1, keep.rownames = TRUE),
setDT(df2, keep.rownames = TRUE)
))[, lapply(.SD, sum, na.rm = TRUE), by = rn]
I prefer the rbind.fill method as you can "merge" > 2 data frames using the same syntax.
This question is related to How can I replace a factor levels with the top n levels (by some metric), plus [other]?. As a metric I want to use the number of occurrences of the factor. I know I can do it by making a list of the occurrences, but I was wondering if there is a prettier way.
Example:
library(data.table);
library(plyr);
fac <- data.table(score = as.factor(c(3,4,5,3,3,3,5)));
ocCnt <- data.table(lapply(fac,count)$score);
fac$occurrence <- 0;
for(i in 1:length(fac$score)){fac$occurrence[i]<-ocCnt[x==fac$score[i]]$freq};
Then I could use the function described in the referenced question/answer:
hotfactor= function(fac,by,n=10,o="other") {
levels(fac)[rank(-xtabs(by~fac))[levels(fac)]>n] <- o
fac
}
To continue the example, if we want only to see the most popular factor we do:
hotfactor(fac$score,fac$occurrence,1);
To get the answer:
[1] 3 other other 3 3 3 other
Levels: 3 other
So my question is, can I do this without having to add a list which counts the occurrences?
Note that I want to do this for the n most popular factors (not just for the most popular factor).
Use table and which.max:
score <- factor(c(3,4,5,3,3,3,5))
levels(score)[- which.max(table(score))] <- "other"
#[1] 3 other other 3 3 3 other
#Levels: 3 other
Obviously this breaks ties by taking the first maximum value.
If you want to keep the top two levels:
score <- factor(c(3, 4,5,3,3,3,5), levels =c(4,3,5))
levels(score)[!levels(score) %in% names(sort(table(score), decreasing = TRUE)[1:2])] <- "other"
#[1] 3 other 5 3 3 3 5
#Levels: other 3 5
If you don't know how many levels you need to group say, 90% of your data and are willing to use dplyr, you could do something along the following lines:
library(dplyr)
df <- data.frame(
f = factor(mapply(rep, letters[1:5], 2^(1:5)) %>% unlist(use.names = F))
)
df %>%
count(f, sort = T) %>%
mutate(p = cumsum(n) / nrow(df))
# A tibble: 5 x 3
# f n p
# <fctr> <int> <dbl>
# 1 e 32 0.5161290
# 2 d 16 0.7741935
# 3 c 8 0.9032258
# 4 b 4 0.9677419
# 5 a 2 1.0000000
(top <- df %>%
count(f, sort = T) %>%
mutate(p = cumsum(n) / nrow(df)) %>%
filter(cumall(p < .91)) %>%
select(f) %>%
unlist(use.names = F))
# [1] e d c
# Levels: a b c d e
levels(df$f) <- factor(c(levels(df$f), 'z'))
df$f[!df$f %in% top] <- 'z'
df %>%
count(f, sort = T) %>%
mutate(p = cumsum(n) / nrow(df))
# A tibble: 4 x 3
# f n p
# <fctr> <int> <dbl>
# 1 e 32 0.5161290
# 2 d 16 0.7741935
# 3 c 8 0.9032258
# 4 z 6 1.0000000