I have the dataframe ds
CountyID ZipCode Value1 Value2 Value3 ... Value25
1 1 0 etc etc etc
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0
and would like to aggregate based on ds$ZipCode and set ds$CountyID equal to the primary county based on the highest ds$Value1. For the above example, it would look like this:
CountyID ZipCode Value1 Value2 Value3 ... Value25
2 1 4 etc etc etc
5 2 2
6 3 3
7 4 9
9 5 1
10 6 0
All the ValueX columns are the sum of that column grouped by ZipCode.
I've tried a bunch of different strategies over the last couple days, but none of them work. The best I've come up with is
#initialize the dataframe
ds_temp = data.frame()
#loop through each subset based on unique zipcodes
for (zip in unique(ds$ZipCode) {
sub <- subset(ds, ds$ZipCode == zip)
len <- length(sub)
maxIndex <- which.max(sub$Value1)
#do the aggregation
row <- aggregate(sub[3:27], FUN=sum, by=list(
CountyID = rep(sub$CountyID[maxIndex], len),
ZipCode = sub$ZipCode))
rbind(ds_temp, row)
}
ds <- ds_temp
I haven't been able to test this on the real data, but with dummy datasets (such as the one above), I keep getting the error "arguments must have the same length). I've messed around with rep() and fixed vectors (eg c(1,2,3,4)) but no matter what I do, the error persists. I also occasionally get an error to the effect of
cannot subset data of type 'closure'.
Any ideas? I've also tried messing around with data.frame(), ddply(), data.table(), dcast(), etc.
You can try this:
data.frame(aggregate(df[,3:27], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
Fully reproducible sample data:
df<-read.table(text="
CountyID ZipCode Value1
1 1 0
2 1 3
3 1 0
4 1 1
5 2 2
6 3 3
7 4 7
8 4 2
9 5 1
10 6 0", header=TRUE)
data.frame(aggregate(df[,3], by=list(df$ZipCode), sum),
CountyID = unlist(lapply(split(df, df$ZipCode),
function(x) x$CountyID[which.max(x$Value1)])))
# Group.1 x CountyID
#1 1 4 2
#2 2 2 5
#3 3 3 6
#4 4 9 7
#5 5 1 9
#6 6 0 10
In response to your comment on Frank's answer, you can preserve the column names by using the formula method in aggregate. Using Franks's data df, this would be
> cbind(aggregate(Value1 ~ ZipCode, df, sum),
CountyID = sapply(split(df, df$ZipCode), function(x) {
with(x, CountyID[Value1 == max(Value1)]) }))
# ZipCode Value1 CountyID
# 1 1 4 2
# 2 2 2 5
# 3 3 3 6
# 4 4 9 7
# 5 5 1 9
# 6 6 0 10
Related
I'm having trouble figuring out how to do the opposite of the answer to this question (and in R not python).
Count the amount of times value A occurs with value B
Basically I have a dataframe with a lot of combinations of pairs of columns like so:
df <- data.frame(id1 = c("1","1","1","1","2","2","2","3","3","4","4"),
id2 = c("2","2","3","4","1","3","4","1","4","2","1"))
I want to count, how often all the values in column A occur in the whole dataframe without the values from column B. So the results for this small example would be the output of:
df_result <- data.frame(id1 = c("1","1","1","2","2","2","3","3","4","4"),
id2 = c("2","3","4","1","3","4","1","4","2","1"),
count = c("4","5","5","3","5","4","2","3","3","3"))
The important criteria for this, is that the final results dataframe is collapsed by the pairs (so in my example rows 1 and 2 are duplicates, and they are collapsed and summed by the total frequency 1 is observed without 2). For tallying the count of occurances, it's important that both columns are examined. I.e. order of columns doesn't matter for calculating the frequency - if column A has 1 and B has 2, this counts the same as if column A has 2 and B has 1.
I can do this very slowly by filtering for each pair, but it's not really feasible for my real data where I have many many different pairs.
Any guidance is greatly appreciated.
First paste the two id columns together to id12 for later matching. Then use sapply to go through all rows to see the records where id1 appears in id12 but id2 doesn't. sum that value and only output the distinct records. Finally, remove the id12 column.
library(dplyr)
df %>% mutate(id12 = paste0(id1, id2),
count = sapply(1:nrow(.),
function(x)
sum(grepl(id1[x], id12) & !grepl(id2[x], id12)))) %>%
distinct() %>%
select(-id12)
Or in base R completely:
id12 <- paste0(df$id1, df$id2)
df$count <- sapply(1:nrow(df), function(x) sum(grepl(df$id1[x], id12) & !grepl(df$id2[x], id12)))
df <- df[!duplicated(df),]
Output
id1 id2 count
1 1 2 4
2 1 3 5
3 1 4 5
4 2 1 3
5 2 3 5
6 2 4 4
7 3 1 2
8 3 4 3
9 4 2 3
10 4 1 3
A full tidyverse version:
library(tidyverse)
df %>%
mutate(id = paste(id1, id2),
count = map(cur_group_rows(), ~ sum(str_detect(id, id1[.x]) & str_detect(id, id2[.x], negate = T))))
A more efficient approach would be to work on a tabulation format:
tab = crossprod(table(rep(seq_len(nrow(df)), ncol(df)), c(df$id1, df$id2)))
#tab
#
# 1 2 3 4
# 1 7 3 2 2
# 2 3 6 1 2
# 3 2 1 4 1
# 4 2 2 1 5
So, now, we have the times each value appears with another (irrespectively of their order in the two columns). Here on, we need a way to subset the above table by each pair and subtract the value of their cooccurence from the value of each id's total appearance.
Make a grid of all combinations:
gr = expand.grid(id1 = colnames(tab), id2 = rownames(tab), stringsAsFactors = FALSE)
Create 2-column matrices to subset the table:
id1.ij = cbind(match(gr$id1, colnames(tab)),
match(gr$id1, rownames(tab)))
id2.ij = cbind(match(gr$id1, colnames(tab)),
match(gr$id2, rownames(tab)))
Subtract the respective values:
cbind(gr, count = tab[id1.ij] - tab[id2.ij])
# id1 id2 count
#1 1 1 0
#2 2 1 3
#3 3 1 2
#4 4 1 3
#5 1 2 4
#6 2 2 0
#7 3 2 3
#8 4 2 3
#9 1 3 5
#10 2 3 5
#11 3 3 0
#12 4 3 4
#13 1 4 5
#14 2 4 4
#15 3 4 3
#16 4 4 0
Of course, if we do not need the full grid of values, we can set:
gr = unique(df)
which results in:
# id1 id2 count
#1 1 2 4
#3 1 3 5
#4 1 4 5
#5 2 1 3
#6 2 3 5
#7 2 4 4
#8 3 1 2
#9 3 4 3
#10 4 2 3
#11 4 1 3
My dataframe looks like this
data = data.frame(ID=c(1,2,3,4,5,6,7,8,9,10),
Gender=c('Male','Female','Female','Female','Male','Female','Male','Male','Female','Female'))
And I have a reference list that looks like this -
ref=list(Male=1,Female=2)
I'd like to replace values in the Gender column using this reference list, without adding a new column to my dataframe.
Here's my attempt
do.call(dplyr::recode, c(list(data), ref))
Which gives me the following error -
no applicable method for 'recode' applied to an object of class
"data.frame"
Any inputs would be greatly appreciated
An option would be do a left_join after stacking the 'ref' list to a two column data.frame
library(dplyr)
left_join(data, stack(ref), by = c('Gender' = 'ind')) %>%
select(ID, Gender = values)
A base R approach would be
unname(unlist(ref)[as.character(data$Gender)])
#[1] 1 2 2 2 1 2 1 1 2 2
In base R:
data$Gender = sapply(data$Gender, function(x) ref[[x]])
You can use factor, i.e.
factor(data$Gender, levels = names(ref), labels = ref)
#[1] 1 2 2 2 1 2 1 1 2 2
You can unlist ref to give you a named vector of codes, and then index this with your data:
transform(data,Gender=unlist(ref)[as.character(Gender)])
ID Gender
1 1 1
2 2 2
3 3 2
4 4 2
5 5 1
6 6 2
7 7 1
8 8 1
9 9 2
10 10 2
Surprisingly, that one works as well:
data$Gender <- ref[as.character(data$Gender)]
#> data
# ID Gender
# 1 1 1
# 2 2 2
# 3 3 2
# 4 4 2
# 5 5 1
# 6 6 2
# 7 7 1
# 8 8 1
# 9 9 2
# 10 10 2
I have a list of events and sequences. I would like to print the sequences in a separate table if event = x is included somewhere in the sequence. See table below:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4
In this case I would like a new table that includes only the sequences where Event=x was included:
Event Sequence
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
Base R solution:
d[d$Sequence %in% d$Sequence[d$Event == "x"], ]
Event Sequence
1: a 1
2: a 1
3: x 1
4: a 3
5: a 3
6: x 3
data.table solution:
library(data.table)
setDT(d)[Sequence %in% Sequence[Event == "x"]]
As you can see syntax/logic is quite similar between these two solutions:
Find event's that are equal to x
Extract their Sequence
Subset table according to specified Sequence
We can use dplyr to group the data and filter the sequence with any "x" in it.
library(dplyr)
df2 <- df %>%
group_by(Sequence) %>%
filter(any(Event %in% "x")) %>%
ungroup()
df2
# A tibble: 6 x 2
Event Sequence
<chr> <int>
1 a 1
2 a 1
3 x 1
4 a 3
5 a 3
6 x 3
DATA
df <- read.table(text = " Event Sequence
1 a 1
2 a 1
3 x 1
4 a 2
5 a 2
6 a 3
7 a 3
8 x 3
9 a 4
10 a 4",
header = TRUE, stringsAsFactors = FALSE)
I need to create (with R) a rolling index of pairs from a data set that includes groups. Consider the following data set:
times <- c(4,3,2)
V1 <- unlist(lapply(times, function(x) seq(1, x)))
df <- data.frame(group = rep(1:length(times), times = times),
V1 = V1,
rolling_index = c(1,1,2,2,3,3,4,5,5))
df
group V1 rolling_index
1 1 1 1
2 1 2 1
3 1 3 2
4 1 4 2
5 2 1 3
6 2 2 3
7 2 3 4
8 3 1 5
9 3 2 5
The data frame I have includes the variables group and V1. Within each group V1 designates a running index (that may or may not start at 1).
I want to create a new indexing variable that looks like rolling_index. This variable groups rows within the same group and consecutive V1 value, thus creating a new rolling index. This new index must be consecutive over groups. If there is an uneven amount of rows within a group (e.g. group 2), then the last, single row gets its own rolling index value.
You can try
library(data.table)
setDT(df)[, gr:=as.numeric(gl(.N, 2, .N)), group][,
rollindex:=cumsum(c(TRUE,abs(diff(gr))>0))][,gr:= NULL]
# group V1 rolling_index rollindex
#1: 1 1 1 1
#2: 1 2 1 1
#3: 1 3 2 2
#4: 1 4 2 2
#5: 2 1 3 3
#6: 2 2 3 3
#7: 2 3 4 4
#8: 3 1 5 5
#9: 3 2 5 5
Or using base R
indx1 <- !duplicated(df$group)
indx2 <- with(df, ave(group, group, FUN=function(x)
gl(length(x), 2, length(x))))
cumsum(c(TRUE,diff(indx2)>0)|indx1)
#[1] 1 1 2 2 3 3 4 5 5
Update
The above methods are based on the 'group' column. Suppose you already have a sequence column ('V1') by group as showed in the example, creation of rolling index is easier
cumsum(!!df$V1 %%2)
#[1] 1 1 2 2 3 3 4 5 5
As mentioned in the post, if the 'V1' column do not start at '1' for some groups, we can get the sequence from the 'group' and then do the cumsum as above
cumsum(!!with(df, ave(seq_along(group), group, FUN=seq_along))%%2)
#[1] 1 1 2 2 3 3 4 5 5
There is probably a simpler way but you can do:
rep_each <- unlist(mapply(function(q,r) {c(rep(2, q),rep(1, r))},
q=table(df$group)%/%2,
r=table(df$group)%%2))
df$rolling_index <- inverse.rle(x=list(lengths=rep_each, values=seq(rep_each)))
df$rolling_index
#[1] 1 1 2 2 3 3 4 5 5
I'm trying to select the column with the highest value for each row in a data.frame. So for instance, the data is set up as such.
> df <- data.frame(one = c(0:6), two = c(6:0))
> df
one two
1 0 6
2 1 5
3 2 4
4 3 3
5 4 2
6 5 1
7 6 0
Then I'd like to set another column based on those rows. The data frame would look like this.
> df
one two rank
1 0 6 2
2 1 5 2
3 2 4 2
4 3 3 3
5 4 2 1
6 5 1 1
7 6 0 1
I imagine there is some sort of way that I can use plyr or sapply here but it's eluding me at the moment.
There might be a more efficient solution, but
ranks <- apply(df, 1, which.max)
ranks[which(df[, 1] == df[, 2])] <- 3
edit: properly spaced!