I have a dataframe df where:
Days Treatment A Treatment B Treatment C
0 5 1 1
1 0 2 3
2 1 1 0
For example, there were 5 individuals receiving Treatment A that survived 0 days and 1 who survived 2, etc. However, I would like it where those 5 individuals now become a unique row, with that cell representing the days they survived:
Patient # A B C
1 0
2 0
3 0
4 0
5 0
6 2
7 0
8 1
9 1
10 2
11 0
12 1
13 1
14 1
Let Patient # = an arbitrary value.
I am sorry if this is not descriptive enough, but I appreciate any and all help you have to offer! I have the dataset in Excel at the moment, but I can place it into R if that's easier.
We can replicate values the 'Days' with each of the 'Patient' column values in a list, then create a list of the sequence, use Map to construct a data.frame and finally use bind_rows
library(dplyr)
lst1 <- lapply(df[-1], function(x) rep(df$Days, x))
bind_rows(Map(function(x, y, z) setNames(data.frame(x, y),
c("Patient", z)), relist(seq_along(unlist(lst1)),
skeleton = lst1), lst1, sub("Treatment\\s+", "", names(lst1))))
-output
# Patient A B C
#1 1 0 NA NA
#2 2 0 NA NA
#3 3 0 NA NA
#4 4 0 NA NA
#5 5 0 NA NA
#6 6 2 NA NA
#7 7 NA 0 NA
#8 8 NA 1 NA
#9 9 NA 1 NA
#10 10 NA 2 NA
#11 11 NA NA 0
#12 12 NA NA 1
#13 13 NA NA 1
#14 14 NA NA 1
Or another option with reshaping into 'long' and then to 'wide'
library(tidyr)
df %>%
pivot_longer(cols = -Days) %>%
separate(name, into = c('name1', 'name2')) %>%
group_by(name2) %>%
summarise(value = rep(Days, value), .groups = 'drop') %>%
mutate(Patient = row_number()) %>%
pivot_wider(names_from = name2, values_from = value)
-output
# A tibble: 14 x 4
# Patient A B C
# <int> <int> <int> <int>
# 1 1 0 NA NA
# 2 2 0 NA NA
# 3 3 0 NA NA
# 4 4 0 NA NA
# 5 5 0 NA NA
# 6 6 2 NA NA
# 7 7 NA 0 NA
# 8 8 NA 1 NA
# 9 9 NA 1 NA
#10 10 NA 2 NA
#11 11 NA NA 0
#12 12 NA NA 1
#13 13 NA NA 1
#14 14 NA NA 1
data
df <- structure(list(Days = 0:2, `Treatment A` = c(5L, 0L, 1L),
`Treatment B` = c(1L,
2L, 1L), `Treatment C` = c(1L, 3L, 0L)), class = "data.frame", row.names = c(NA,
-3L))
Related
I want to group my data in different chunks when the data is continuous. Trying to get the group column from dummy data like this:
a b group
<dbl> <dbl> <dbl>
1 1 1 1
2 2 2 1
3 3 3 1
4 4 NA NA
5 5 NA NA
6 6 NA NA
7 7 12 2
8 8 15 2
9 9 NA NA
10 10 25 3
I tried using
test %>% mutate(test = complete.cases(.)) %>%
group_by(group = cumsum(test == TRUE)) %>%
select(group, everything())
But it doesn't work as expected:
group a b test
<int> <dbl> <dbl> <lgl>
1 1 1 1 TRUE
2 2 2 2 TRUE
3 3 3 3 TRUE
4 3 4 NA FALSE
5 3 5 NA FALSE
6 3 6 NA FALSE
7 4 7 12 TRUE
8 5 8 15 TRUE
9 5 9 NA FALSE
10 6 10 25 TRUE
Any advice?
Using rle in base R -
transform(df, group1 = with(rle(!is.na(b)), rep(cumsum(values), lengths))) |>
transform(group1 = replace(group1, is.na(b), NA))
# a b group group1
#1 1 1 1 1
#2 2 2 1 1
#3 3 3 1 1
#4 4 NA NA NA
#5 5 NA NA NA
#6 6 NA NA NA
#7 7 12 2 2
#8 8 15 2 2
#9 9 NA NA NA
#10 10 25 3 3
A couple of approaches to consider if you wish to use dplyr for this.
First, you could look at transition from non-complete cases (using lag) to complete cases.
library(dplyr)
test %>%
mutate(test = complete.cases(.)) %>%
group_by(group = cumsum(test & !lag(test, default = F))) %>%
mutate(group = replace(group, !test, NA))
Alternatively, you could add row numbers to your data.frame. Then, you could filter to include only complete cases, and group_by enumerating with cumsum based on gaps in row numbers. Then, join back to original data.
test$rn <- seq.int(nrow(test))
test %>%
filter(complete.cases(.)) %>%
group_by(group = c(0, cumsum(diff(rn) > 1)) + 1) %>%
right_join(test) %>%
arrange(rn) %>%
dplyr::select(-rn)
Output
a b group
<int> <int> <dbl>
1 1 1 1
2 2 2 1
3 3 3 1
4 4 NA NA
5 5 NA NA
6 6 NA NA
7 7 12 2
8 8 15 2
9 9 NA NA
10 10 25 3
Using data.table, get rleid then remove group IDs for NAs, then fix the sequence with factor to integer conversion:
library(data.table)
setDT(test)[, group1 := {
x <- complete.cases(test)
grp <- rleid(x)
grp[ !x ] <- NA
as.integer(factor(grp))
}]
# a b group group1
# 1: 1 1 1 1
# 2: 2 2 1 1
# 3: 3 3 1 1
# 4: 4 NA NA NA
# 5: 5 NA NA NA
# 6: 6 NA NA NA
# 7: 7 12 2 2
# 8: 8 15 2 2
# 9: 9 NA NA NA
# 10: 10 25 3 3
I am trying to sum the row of values if any column have values but not working for me like below
df=data.frame(
x3=c(2,NA,3,5,4,6,NA,NA,3,3),
x4=c(0,NA,NA,6,5,6,NA,0,4,2))
df$summ <- ifelse(is.na(c(df[,"x3"] & df[,"x4"])),NA,rowSums(df[,c("x3","x4")], na.rm=TRUE))
the output should be like
An alternative solution:
library(data.table)
setDT(df)[!( is.na(x3) & is.na(x4)),summ:=rowSums(.SD, na.rm = T)]
You can do :
df <- transform(df, summ = ifelse(is.na(x3) & is.na(x4), NA,
rowSums(df, na.rm = TRUE)))
df
# x3 x4 summ
#1 2 0 2
#2 NA NA NA
#3 3 NA 3
#4 5 6 11
#5 4 5 9
#6 6 6 12
#7 NA NA NA
#8 NA 0 0
#9 3 4 7
#10 3 2 5
In general for any number of columns :
cols <- c('x3', 'x4')
df <- transform(df, summ = ifelse(rowSums(is.na(df[cols])) == length(cols),
NA, rowSums(df, na.rm = TRUE)))
Try the code below with rowSums + replace
df$summ <- replace(rowSums(df, na.rm = TRUE), rowSums(is.na(df)) == 2, NA)
which gives
> df
x3 x4 summ
1 2 0 2
2 NA NA NA
3 3 NA 3
4 5 6 11
5 4 5 9
6 6 6 12
7 NA NA NA
8 NA 0 0
9 3 4 7
10 3 2 5
This is not much different from already posted answers, however, it contains some useful functions:
library(dplyr)
df %>%
rowwise() %>%
mutate(Count = ifelse(all(is.na(cur_data())), NA,
sum(c_across(everything()), na.rm = TRUE)))
# A tibble: 10 x 3
# Rowwise:
x3 x4 Count
<dbl> <dbl> <dbl>
1 2 0 2
2 NA NA NA
3 3 NA 3
4 5 6 11
5 4 5 9
6 6 6 12
7 NA NA NA
8 NA 0 0
9 3 4 7
10 3 2 5
Hello a really simple question but I have just got stuck, how do I add a conditional column containing number 1 where completed column is not NA?
id completed
<chr> <chr>
1 abc123sdf 35929
2 124cv NA
3 125xvdf 36295
4 126v NA
5 127sdsd 43933
6 128dfgs NA
7 129vsd NA
8 130sdf NA
9 131sdf NA
10 123sdfd NA
I need this to calculate an overall percent of completed column/id.
(Additional question - how can I do this in dplyr without using a helper column?)
Thanks
You can use is.na to check for NA values.
library(dplyr)
df %>% mutate(newcol = as.integer(!is.na(completed)))
# id completed newcol
#1 abc123sdf 35929 1
#2 124cv NA 0
#3 125xvdf 36295 1
#4 126v NA 0
#5 127sdsd 43933 1
#6 128dfgs NA 0
#7 129vsd NA 0
#8 130sdf NA 0
#9 131sdf NA 0
#10 123sdfd NA 0
library("dplyr")
df <- data.frame(id = 1:10,
completed = c(35929, NA, 36295, NA, 43933, NA, NA, NA, NA, NA))
df %>%
mutate(is_na = as.integer(!is.na(completed)))
#> id completed is_na
#> 1 1 35929 1
#> 2 2 NA 0
#> 3 3 36295 1
#> 4 4 NA 0
#> 5 5 43933 1
#> 6 6 NA 0
#> 7 7 NA 0
#> 8 8 NA 0
#> 9 9 NA 0
#> 10 10 NA 0
But you shouldn't need this extra column to calculate a percentage, you can just use na.rm:
df %>%
mutate(pct = completed / sum(completed, na.rm = TRUE))
#> id completed pct
#> 1 1 35929 0.3093141
#> 2 2 NA NA
#> 3 3 36295 0.3124650
#> 4 4 NA NA
#> 5 5 43933 0.3782209
#> 6 6 NA NA
#> 7 7 NA NA
#> 8 8 NA NA
#> 9 9 NA NA
#> 10 10 NA NA
We can also do
library(dplyr)
df %>%
mutate(newcol = +(!is.na(completed)))
How to efficiently update multiple columns from NA to 0? Don't want to update all NA to 0. Only certain columns need be updated.
My current solution: Is there a better method?
dataframe$col1 = replace(dataframe$col1, is.na(dataframe$col1), 0)
dataframe$col2 = replace(dataframe$col2, is.na(dataframe$col2), 0)
dataframe$col3 = replace(dataframe$col3, is.na(dataframe$col3), 0)
Syntax used and this is not working as expected. Meaning, does not replace NA to 0.
dataframe = dataframe %>% mutate(across(c('col1', 'col2', 'col3'), ~ replace(., all(is.na(.)), 0)))
Sample Data.
structure(list(col1 = c(63755.4062, 61131.3242,
61131.3242, 192055.25, 191429.9844, 190076.4688), col2 = c(18.8754,
14.6002, 14.6002, 24.0053, 24.4012, 25.3588), col3 = c(NA, NA, NA, 45.6442, 43.9821, 47.2581)), row.names = c(NA, 6L), class = "data.frame")
Following worked. Thanks #MATT, and #Karthik.
dataframe = dataframe %>% mutate(across(c('col1', 'col2', 'col3'), ~ ~tidyr::replace_na(., 0)))
I'm not sure why Karthik's solution still returns NA for your sample data, but using replace_na from tidyr seems to work:
library(tidyr)
dataframe %>% mutate(across(c('col1', 'col2', 'col3'), ~ replace_na(., 0)))
Which gives us:
col1 col2 col3
1 63755.41 18.8754 0.0000
2 61131.32 14.6002 0.0000
3 61131.32 14.6002 0.0000
4 192055.25 24.0053 45.6442
5 191429.98 24.4012 43.9821
6 190076.47 25.3588 47.2581
Does this work:
library(dplyr)
library(tibble)
df <- tibble(c1 = round(rnorm(10, 10,1)),
c2 = NA_real_,
c3 = round(rnorm(10, 10,1)),
c4 = NA_real_,
c5 = round(rnorm(10, 10,1)),
c6 = NA_real_)
df
# A tibble: 10 x 6
c1 c2 c3 c4 c5 c6
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 12 NA 11 NA 11 NA
2 9 NA 10 NA 10 NA
3 11 NA 11 NA 10 NA
4 11 NA 9 NA 10 NA
5 10 NA 9 NA 10 NA
6 9 NA 13 NA 12 NA
7 10 NA 10 NA 9 NA
8 10 NA 10 NA 9 NA
9 11 NA 11 NA 10 NA
10 10 NA 10 NA 10 NA
df %>% mutate(across(c3:c6, ~ replace(., all(is.na(.)), 0)))
# A tibble: 10 x 6
c1 c2 c3 c4 c5 c6
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 12 NA 11 0 11 0
2 9 NA 10 0 10 0
3 11 NA 11 0 10 0
4 11 NA 9 0 10 0
5 10 NA 9 0 10 0
6 9 NA 13 0 12 0
7 10 NA 10 0 9 0
8 10 NA 10 0 9 0
9 11 NA 11 0 10 0
10 10 NA 10 0 10 0
The quickest way IMO would be to subset the columns you want to edit as a new dataframe, edit all NAs in the subset to 0, then overwrite your original df's selected columns.
DFsubset <- DF[,10:12] #whichever columns
DFsubset[is.na(DFsubset) == T] <- 0
DF[,10:12] <- DFsubset
I have a dataframe in R:
Subject T O E P Score
1 0 1 0 1 256
2 1 0 1 0 325
2 0 1 0 1 125
3 0 1 0 1 27
4 0 0 0 1 87
5 0 1 0 1 125
6 0 1 1 1 100
This is just a display of the dataframe. In reality, I have a lot of lines for each of the subjects. But the subjects are only from 1 to 6
For each Subject, the possible values are:
T : 0 or 1
O : 0 or 1
E : 0 or 1
P : 0 or 1
Score : Numeric value
I want to create a new dataframe with 6 lines (one for each subject) and the calculated MEAN score for each of these combinations :
T , O , E , P , TO , TE, TP, OE , OP , PE , TOP , TOE , POE , PET
The above will the columns of the new dataframe.
The final output should look like this
Subject T O E P TO TE TP OE OP PE TOP TOE POE PET
1
2
3
4
5
6
For each of these lines x columns the value is the MEAN SCORE
I tried aggregate and table but I can't seem to get what I want
Sorry I am new to R
Thanks
I had to rebuild sample data to answer the question as I understood it, tell me if it works for you :
set.seed(2)
df <- data.frame(subject=sample(1:3,9,T),
T = sample(c(0,1),9,T),
O = sample(c(0,1),9,T),
E = sample(c(0,1),9,T),
P = sample(c(0,1),9,T),
score=round(rnorm(9,10,3)))
# subject T O E P score
# 1 1 1 0 0 1 12
# 2 3 1 0 1 0 9
# 3 2 0 1 0 1 13
# 4 1 1 0 0 0 3
# 5 3 0 1 0 1 14
# 6 3 0 0 1 0 13
# 7 1 1 0 1 0 17
# 8 3 1 0 1 0 12
# 9 2 0 0 1 1 14
cols1 <- c("T","O","E","P")
df$comb <- apply(df[cols1],1,function(x) paste(names(df[cols1])[as.logical(x)],collapse=""))
# subject T O E P score comb
# 1 1 1 0 0 1 12 TP
# 2 3 1 0 1 0 9 TE
# 3 2 0 1 0 1 13 OP
# 4 1 1 0 0 0 3 T
# 5 3 0 1 0 1 14 OP
# 6 3 0 0 1 0 13 E
# 7 1 1 0 1 0 17 TE
# 8 3 1 0 1 0 12 TE
# 9 2 0 0 1 1 14 EP
library(tidyverse)
df %>%
group_by(subject,comb) %>%
summarize(score=mean(score)) %>%
spread(comb,score) %>%
ungroup
# # A tibble: 3 x 7
# subject E EP OP T TE TP
# * <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 1 NA NA NA 3 17.0 12
# 2 2 NA 14 13 NA NA NA
# 3 3 13 NA 14 NA 10.5 NA
The second step in base R:
means <- aggregate(score ~ subject + comb,df,mean)
means2 <- reshape(means,timevar="comb",idvar="subject",direction="wide")
setNames(means2,c("subject",sort(unique(df$comb))))
# subject E EP OP T TE TP
# 1 3 13 NA 14 NA 10.5 NA
# 2 2 NA 14 13 NA NA NA
# 5 1 NA NA NA 3 17.0 12
I'd do it like this:
# using your table data
df = read.table(text =
"Subject T O E P Score
1 0 1 0 1 256
2 1 0 1 0 325
2 0 1 0 1 125
3 0 1 0 1 27
4 0 0 0 1 87
5 0 1 0 1 125
6 0 1 1 1 100", stringsAsFactors = FALSE, header=TRUE)
# your desired column names
new_names <- c("T", "O", "E", "P", "TO", "TE", "TP", "OE",
"OP", "PE", "TOP", "TOE", "POE", "PET")
# assigning each of your scores to one of the desired column names
assign_comb <- function(dfrow) {
selection <- c("T", "O", "E", "P")[as.logical(dfrow[2:5])]
do.call(paste, as.list(c(selection, sep = "")))
}
df$comb <- apply(df, 1, assign_comb)
# aggregate all the means together
df_agg <- aggregate(df$Score ~ df$comb + df$Subject, FUN = mean)
# reshape the data to wide format
df_new <- reshape(df_agg, v.names = "df$Score", idvar = "df$Subject",
timevar = "df$comb", direction = "wide")
# clean up the column names to match your desired output
# any column names not found will be added as NA
colnames(df_new) <- gsub("df\\$|Score\\.", "", colnames(df_new))
df_new[, new_names[!new_names %in% colnames(df_new)]] <- NA
df_new <- df_new[, c("Subject", new_names)]
With the result:
> df_new
Subject T O E P TO TE TP OE OP PE TOP TOE POE PET
1 1 NA NA NA NA NA NA NA NA 256 NA NA NA NA NA
2 2 NA NA NA NA NA 325 NA NA 125 NA NA NA NA NA
4 3 NA NA NA NA NA NA NA NA 27 NA NA NA NA NA
5 4 NA NA NA 87 NA NA NA NA NA NA NA NA NA NA
6 5 NA NA NA NA NA NA NA NA 125 NA NA NA NA NA
7 6 NA NA NA NA NA NA NA NA NA NA NA NA NA NA