There is a injury score called ISS score
I have a table of injury data in rows according to pt ID.
I would like to obtain the top three values for the 6 injury columns.
Column values range from 0-5.
pt_id head face abdo pelvis Extremity External
1 4 0 0 1 0 3
2 3 3 5 0 3 2
3 0 0 2 1 1 1
4 2 0 0 0 0 1
5 5 0 0 2 0 1
My output for the above example would be
pt-id n1 n2 n3
1 4 3 1
2 5 3 3
3 2 1 1
4 2 1 0
5 5 2 1
values can be in a list or in new columns as calculating the score is simple from that point on.
I had thought that I would be able to create a list for the 6 injury columns and then apply a sort to each list taking the top three values. My code for that was:
ais$ais_list <- setNames(split(ais[,2:7], seq(nrow(ais))), rownames(ais))
But I struggled to apply the sort to the lists within the data frame as unfortunately some of the data in my data set includes NA values
We could use apply row-wise and sort the dataframe and take only first three values in each row.
cbind(df[1], t(apply(df[-1], 1, sort, decreasing = TRUE)[1:3, ]))
# pt_id 1 2 3
#1 1 4 3 1
#2 2 5 3 3
#3 3 2 1 1
#4 4 2 1 0
#5 5 5 2 1
As some values may contain NA it is better we apply sort using anonymous function and then take take top 3 values using head.
cbind(df[1], t(apply(df[-1], 1, function(x) head(sort(x, decreasing = TRUE), 3))))
A tidyverse option is to first gather the data, arrange it in descending order and for every row select only first three values. We then replace the injury column with the column names which we want and finally spread the data back to wide format.
library(tidyverse)
df %>%
gather(injury, value, -pt_id) %>%
arrange(desc(value)) %>%
group_by(pt_id) %>%
slice(1:3) %>%
mutate(injury = 1:3) %>%
spread(injury, value)
# pt_id `1` `2` `3`
# <int> <int> <int> <int>
#1 1 4 3 1
#2 2 5 3 3
#3 3 2 1 1
#4 4 2 1 0
#5 5 5 2 1
Related
I was was if there is a way to rank-order rows of my Data below such that rows that simultaneously have the largest values on each of risk1, risk2 and risk3 (NOT TOTAL Of the three) are at the top?
For example, in my Desired_output, you see that id == 4 simultaneously has the largest values on risk1, risk2 and risk3 (4,3,2).
For all other ids, there is a 1 or 0 on at least one of the risk1, risk2 and risk3.
Note: Tie's are fine. 4,3,2 == 2,3,4 == 3,2,4.
Data = data.frame(id=1:4,risk1 = c(1,3,5,4), risk2 = c(8,2,1,3), risk3 = c(0,1,4,2))
Desired_output = read.table(h=T,text="
id risk1 risk2 risk3
4 4 3 2
3 5 1 4
2 3 2 1
1 1 8 0
")
Maybe this helps - loop over the rows, sort the elements, paste, convert to numeric, use that to order the rows
Data[order(-apply(Data[-1], 1, \(x)
as.numeric(paste(sort(x), collapse = "")))),]
-output
id risk1 risk2 risk3
4 4 4 3 2
3 3 5 1 4
2 2 3 2 1
1 1 1 8 0
This does the trick:
library(dplyr)
Data %>%
arrange(-row_number())
id risk1 risk2 risk3
1 4 4 3 2
2 3 5 1 4
3 2 3 2 1
4 1 1 8 0
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
This question already has answers here:
How to create a consecutive group number
(13 answers)
Closed 2 years ago.
I have data from an experiment that has multiple rows per item (each row has the reading time for one word of a sentence of n words), and multiple items per subject. Items can be varying numbers of rows. Items were presented in a random order, and their order in the data as initially read in reflects the sequence they saw the items in. What I'd like to do is add a column that contains the order in which the subject saw that item (i.e., 1 for the first item, 2 for the second, etc.).
Here's an example of some input data that has the relevant properties:
d <- data.frame(Subject = c(1,1,1,1,1,2,2,2,2,2),
Item = c(2,2,2,1,1,1,1,2,2,2))
Subject Item
1 2
1 2
1 2
1 1
1 1
2 1
2 1
2 2
2 2
2 2
And here's the output I want:
Subject Item order
1 2 1
1 2 1
1 2 1
1 1 2
1 1 2
2 1 1
2 1 1
2 2 2
2 2 2
2 2 2
I know I can do this by setting up a temp data frame that filters d to unique combinations of Subject and Item, adding order to that as something like 1:n() or row_number(), and then using a join function to put it back together with the main data frame. What I'd like to know is whether there's a way to do this without having to create a new data frame just to store the order---can this be done inside dplyr's mutate somehow if I group by Subject and Item, for instance?
Here's one way:
d %>%
group_by(Subject) %>%
mutate(order = match(Item, unique(Item))) %>%
ungroup()
# # A tibble: 10 x 3
# Subject Item order
# <dbl> <dbl> <int>
# 1 1 2 1
# 2 1 2 1
# 3 1 2 1
# 4 1 1 2
# 5 1 1 2
# 6 2 1 1
# 7 2 1 1
# 8 2 2 2
# 9 2 2 2
# 10 2 2 2
Here is a base R option
transform(d,
order = ave(Item, Subject, FUN = function(x) as.integer(factor(x, levels = unique(x))))
)
or
transform(d,
order = ave(Item, Subject, FUN = function(x) match(x, unique(x)))
)
both giving
Subject Item order
1 1 2 1
2 1 2 1
3 1 2 1
4 1 1 2
5 1 1 2
6 2 1 1
7 2 1 1
8 2 2 2
9 2 2 2
10 2 2 2
I have 500 datasets (panel data). In each I have a time series (week) across different shops (store). Within each shop, I would need to add missing time series observations.
A sample of my data would be:
store week value
1 1 50
1 3 52
1 4 10
2 1 4
2 4 84
2 5 2
which I would like to look like:
store week value
1 1 50
1 2 0
1 3 52
1 4 10
2 1 4
2 2 0
2 3 0
2 4 84
2 5 2
I currently use the following code (which works, but takes very very long on my data):
stores<-unique(mydata$store)
for (i in 1:length(stores)){
mydata <- merge(
expand.grid(week=min(mydata$week):max(mydata$week)),
mydata, all=TRUE)
mydata[is.na(mydata)] <- 0
}
Are there better and more efficient ways to do so?
Here's a dplyr/tidyr option you could try:
library(dplyr); library(tidyr)
group_by(df, store) %>%
complete(week = full_seq(week, 1L), fill = list(value = 0))
#Source: local data frame [9 x 3]
#
# store week value
# (int) (int) (dbl)
#1 1 1 50
#2 1 2 0
#3 1 3 52
#4 1 4 10
#5 2 1 4
#6 2 2 0
#7 2 3 0
#8 2 4 84
#9 2 5 2
By default, if you don't specify the fill parameter, new rows will be filled with NA. Since you seem to have many other columns, I would advise to leave out the fill parameter so you end up with NAs, and if required, make another step with mutate_each to turn NAs into 0 (if that's appropriate).
group_by(df, store) %>%
complete(week = full_seq(week, 1L)) %>%
mutate_each(funs(replace(., which(is.na(.)), 0)), -store, -week)
I've checked this issue but couldn't find a matching entry.
Say you have 2 DFs:
df1:mode df2:sex
1 1
2 2
3
And a DF3 where most of the combinations are not present, e.g.
mode | sex | cases
1 1 9
1 1 2
2 2 7
3 1 2
1 2 5
and you want to summarise it with dplyr obtaining all combinations (with not existent ones=0):
mode | sex | cases
1 1 11
1 2 5
2 1 0
2 2 7
3 1 2
3 2 0
If you do a single left_join (left_join(df1,df3) you recover the modes not in df3, but 'Sex' appears as 'NA', and the same if you do left_join(df2,df3).
So how can you do both left join to recover all absent combinations, with cases=0? dplyr preferred, but sqldf an option.
Thanks in advance, p.
The development version of tidyr, tidyr_0.2.0.9000, has a new function called complete that I saw the other day that seems like it was made for just this sort of situation.
The help page says:
This is a wrapper around expand(), left_join() and replace_na that's
useful for completing missing combinations of data. It turns
implicitly missing values into explicitly missing values.
To add the missing combinations of df3 and fill with 0 values instead, you would do:
library(tidyr)
library(dplyr)
df3 %>% complete(mode, sex, fill = list(cases = 0))
mode sex cases
1 1 1 9
2 1 1 2
3 1 2 5
4 2 1 0
5 2 2 7
6 3 1 2
7 3 2 0
You would still need to group_by and summarise to get the final output you want.
df3 %>% complete(mode, sex, fill = list(cases = 0)) %>%
group_by(mode, sex) %>%
summarise(cases = sum(cases))
Source: local data frame [6 x 3]
Groups: mode
mode sex cases
1 1 1 11
2 1 2 5
3 2 1 0
4 2 2 7
5 3 1 2
6 3 2 0
First here's you data in a more friendly, reproducible format
df1 <- data.frame(mode=1:3)
df2 <- data.frame(sex=1:2)
df3 <- data.frame(mode=c(1,1,2,3,1), sex=c(1,1,2,1,2), cases=c(9,2,7,2,5))
I don't see an option for a full outer join in dplyr, so I'm going to use base R here to merge df1 and df2 to get all mode/sex combinations. Then i left join that to the data and replace NA values with zero.
mm <- merge(df1,df2) %>% left_join(df3)
mm$cases[is.na(mm$cases)] <- 0
mm %>% group_by(mode,sex) %>% summarize(cases=sum(cases))
which gives
mode sex cases
1 1 1 11
2 1 2 5
3 2 1 0
4 2 2 7
5 3 1 2
6 3 2 0