I have a data frame where each row is an observation and I have two columns:
the group membership of the observation
the outcome for the observation.
I'm trying to create a new variable outcome_change that takes a value of 1 if outcome is NOT identical for all observations in a given group and 0 otherwise.
Shown in the below code (dat) is an example of the data I have. Meanwhile, dat_out1 shows what I'm looking for the code to produce in the presence of no NA values. The dat_out2 is identical except it shows that the same results arise when there are missing values in a group's values.
Surely there is somewhat to do this with dplyr::group_by()? I don't know how to make these comparisons within groups.
# Input (2 groups: 1 with identical values of outcome
# in the group (group a) and 1 with differing values of
# outcome in the group (group b)
dat <- data.frame(group = c("a","a","a","b","b","b"),
outcome = c(1,1,1,3,2,2))
# Output 1: add a variable for all observations belonging to
# a group where the outcome changed within each group
dat_out1 <- data.frame(group = c("a","a","a","b","b","b"),
outcome = c(1,1,1,3,2,2),
outcome_change = c(0,0,0,1,1,1))
# Output 2: same as Output 1, but able to ignore NA values
dat_out2 <- data.frame(group = c("a","a","a","b","b","b"),
outcome = c(1,1,NA,3,2,NA),
outcome_change = c(0,0,0,1,1,1))
Here is an aproach:
library(tidyverse)
dat %>%
group_by(group) %>%
mutate(outcome_change = ifelse(length(unique(outcome[!is.na(outcome)])) > 1, 1, 0))
#output
# A tibble: 6 x 3
# Groups: group [2]
group outcome outcome_change
<fctr> <dbl> <dbl>
1 a 1 0
2 a 1 0
3 a 1 0
4 b 3 1
5 b 2 1
6 b 2 1
with dat2
# A tibble: 6 x 3
# Groups: group [2]
group outcome outcome_change
<fctr> <dbl> <dbl>
1 a 1 0
2 a 1 0
3 a NA 0
4 b 3 1
5 b 2 1
6 b NA 1
library(dplyr)
dat <- data.frame(group = c("a","a","a","b","b","b"),
outcome = c(1,1,1,3,2,2))
dat2 <- data.frame(group = c("a","a","a","b","b","b"),
outcome = c(1,1,NA,3,2,NA))
dat_out1 <- dat %>% group_by(group) %>%
mutate(outcome_change = ifelse(min(outcome) == max(outcome), 0, 1))
dat_out2 <- dat2 %>% group_by(group) %>%
mutate(outcome_change = ifelse(min(outcome, na.rm = TRUE) == max(outcome, na.rm = TRUE), 0, 1))
Here is an option using data.table
library(data.table)
setDT(dat1)[, outcome_change := as.integer(uniqueN(outcome[!is.na(outcome)])>1), group]
dat1
# group outcome outcome_change
#1: a 1 0
#2: a 1 0
#3: a 1 0
#4: b 3 1
#5: b 2 1
#6: b 2 1
If we apply the same with 'dat2'
dat2
# group outcome outcome_change2
#1: a 1 0
#2: a 1 0
#3: a NA 0
#4: b 3 1
#5: b 2 1
#6: b NA 1
Related
I have two dataframes df1 and df2. They both have a column 'ID'. For each row in DF1, I would like to find out how many duplicates of its ID there are in df2 and add the count to that row. If there are no duplicates, the count should return as 0.
# # A tibble: 4 x 3
# ID a b
# <dbl> <dbl> <dbl>
# 1 1_234 1 1
# 2 1_235 1 2
# 3 2_222 1 1
# 4 2_654 1 2
# # A tibble: 4 x 3
# ID a b
# <dbl> <dbl> <dbl>
# 1 1_234 1 1
# 2 1_235 1 2
# 3 1_234 1 1
# 4 3_234 1 2
Using dplyr:
Your data:
df1 <- data.frame(ID = c("1_234","1_235","2_222","2_654"),
a = c(1,1,1,1),
b = c(1,2,1,2))
df2 <- data.frame(ID = c("1_234","1_235","1_234","3_235"),
a = c(1,1,1,1),
b = c(1,2,1,2))
Edit: considering only the IDs:
output <- left_join(df1,
as.data.frame(table(df2$ID)),
by = c("ID" = "Var1")) %>%
mutate(Freq = ifelse(is.na(Freq), 0, Freq))
Output:
ID a b Freq
1 1_234 1 1 2
2 1_235 1 2 1
3 2_222 1 1 0
4 2_654 1 2 0
A base R option using subset + aggregate
subset(
aggregate(
n ~ .,
rbind(
cbind(df1, n = 1),
cbind(df2, n = 1)
), function(x) length(x) - 1
), ID %in% df1$ID
)
gives
ID a b n
1 1_234 1 1 2
2 2_222 1 1 0
3 1_235 1 2 1
4 2_654 1 2 0
I think you can do it with a simple sapply() and base r (no extra packages).
df1$count <- sapply(df1$ID, function(x) sum(df2$ID == x))
We may also use outer
df1$count <- rowSums(outer(df1$ID, df2$ID, FUN = `==`))
df1$count
[1] 2 1 0 0
We could use semi_join and n() to get the count of duplicates:
library(dplyr)
df1 %>%
semi_join(df2, by="ID") %>%
summarise(duplicates_df1_df2 = n())
Output:
duplicates_df1_df2
1 2
I have a data frame with a group, a condition that differs by group, and an index within each group:
df <- data.frame(group = c(rep(c("A", "B", "C"), each = 3)),
condition = rep(c(0,1,1), each = 3),
index = c(1:3,1:3,2:4))
> df
group condition index
1 A 0 1
2 A 0 2
3 A 0 3
4 B 1 1
5 B 1 2
6 B 1 3
7 C 1 2
8 C 1 3
9 C 1 4
I would like to slice the data within each group, filtering out all but the row with the lowest index. However, this filter should only be applied when the condition applies, i.e., condition == 1. My solution was to compute a ranking on the index within each group and filter on the combination of condition and rank:
df %>%
group_by(group) %>%
mutate(rank = order(index)) %>%
filter(case_when(condition == 0 ~ TRUE,
condition == 1 & rank == 1 ~ TRUE))
# A tibble: 5 x 4
# Groups: group [3]
group condition index rank
<chr> <dbl> <int> <int>
1 A 0 1 1
2 A 0 2 2
3 A 0 3 3
4 B 1 1 1
5 C 1 2 1
This left me wondering whether there is a faster solution that does not require a separate ranking variable, and potentially uses slice_min() instead.
You can use filter() to keep all cases where the condition is zero or the index equals the minimum index.
library(dplyr)
df %>%
group_by(group) %>%
filter(condition == 0 | index == min(index))
# A tibble: 5 x 3
# Groups: group [3]
group condition index
<chr> <dbl> <int>
1 A 0 1
2 A 0 2
3 A 0 3
4 B 1 1
5 C 1 2
An option with slice
library(dplyr)
df %>%
group_by(group) %>%
slice(unique(c(which(condition == 0), which.min(index))))
How can I create a new data frame with the smallest values group by a column.
For example this df:
df <- read.table(header = TRUE, text = 'Gene Value
A 12
A 10
B 3
B 0
B 6
C 1
D 0
D 4')
Now with:
test <- setDT(df)[, .SD[which.min(Value)], by=Gene]
I get this:
> test
Gene Value
1: A 10
2: B 0
3: C 1
4: D 0
But how can I use a second condition for Value > 0 here? I want to have this output:
> test
Gene Value
1: A 10
2: B 3
3: C 1
4: D 4
Could do:
setDT(df)[, .(Value = min(Value[Value > 0])), by=Gene]
Output:
Gene Value
1: A 10
2: B 3
3: C 1
4: D 4
Using tidyverse you can group, filter and then summarize the min value:
library(tidyverse)
df2 <- df %>%
group_by(Gene) %>%
filter(Value != 0) %>%
summarise(Value = min(Value))
# A tibble: 4 x 2
Gene Value
<fct> <dbl>
1 A 10
2 B 3
3 C 1
4 D 4
Using aggregate from base R
aggregate(Value ~ Gene, subset(df, Value > 0), min)
# Gene Value
#1 A 10
#2 B 3
#3 C 1
#4 D 4
Slightly difficult to phrase, as far as I saw none of the similar questions answered my problem.
I have a data.frame such as:
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
df1
id val
1 a NA
2 a NA
3 a NA
4 a NA
5 b 1
6 b 2
7 b 2
8 b 3
9 c NA
10 c 2
11 c NA
12 c 3
and I want to get rid of all the NA values (easy enough using e.g. filter() ) but make sure that if this removes all of one id value (in this case it removes every instance of "a") that one extra row is inserted of (e.g.) a = 0
so that:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c 2
7 c 3
obviously easy enough to do this in a roundabout way but I was wondering if there's a tidy/elegant way to do this. I thought tidyr::complete() might help but not entirely sure how to apply it to a case like this
I don't care about the order of the rows
Cheers!
edit: updated with clearer desired output. might make desired answers submitted before that a bit less clear
Another idea using dplyr,
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(val = ifelse(row_number() == 1 & all(is.na(val)), 0, val)) %>%
na.omit()
which gives,
# A tibble: 5 x 2
# Groups: id [2]
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
We may do
df1 %>% group_by(id) %>% do(if(all(is.na(.$val))) replace(.[1, ], 2, 0) else na.omit(.))
# A tibble: 5 x 2
# Groups: id [2]
# id val
# <fct> <dbl>
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
After grouping by id, if everything in val is NA, then we leave only the first row with the second element replaced by 0, otherwise the same data is returned after applying na.omit.
In a more readable format that would be
df1 %>% group_by(id) %>%
do(if(all(is.na(.$val))) data.frame(id = .$id[1], val = 0) else na.omit(.))
(Here I presume that you indeed want to get rid of all NA values; otherwise there is no need for na.omit.)
df1[is.na(df1)] <- 0
df1[!(duplicated(df1$id) & df1$val == 0), ]
id val
1 a 0
5 b 1
6 b 2
7 b 2
8 b 3
Base R option is to find groups with all NAs and transform them by changing their val to 0 and select only unique rows so that there is only one row per group. We rbind this dataframe with the groups which are !all_NA.
all_NA <- with(df1, ave(is.na(val), id, FUN = all))
rbind(unique(transform(df1[all_NA, ], val = 0)), df1[!all_NA, ])
# id val
#1 a 0
#5 b 1
#6 b 2
#7 b 2
#8 b 3
dplyr option looks ugly but one way is to make two groups of dataframes one with groups of all NA values and other with groups of all non-NA values. For groups with all NA values we add row with it's id and val as 0 and bind this to the other group.
library(dplyr)
bind_rows(df1 %>%
group_by(id) %>%
filter(all(!is.na(val))),
df1 %>%
group_by(id) %>%
filter(all(is.na(val))) %>%
ungroup() %>%
summarise(id = unique(id),
val = 0)) %>%
arrange(id)
# id val
# <fct> <dbl>
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Changed the df to make example more exhaustive -
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
library(dplyr)
df1 %>%
group_by(id) %>%
mutate(case=sum(is.na(val))==n(), row_num=row_number() ) %>%
mutate(val=ifelse(is.na(val)&case,0,val)) %>%
filter( !(case&row_num!=1) ) %>%
select(id, val)
Output
id val
<fct> <dbl>
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
Another base approach, one that doesn't maintain the order of the rows and takes advantage of factors remembering lost values:
df1 <- na.omit(df1)
df1 <- rbind(
df1,
data.frame(
id = levels(df1$id)[!levels(df1$id) %in% df1$id],
val = 0)
)
I do personally prefer the dplyr approach given by Sotos, as I don't like rbind-ing data.frames back together so it's a matter of taste, but this isn't unbearably complicated by my eye. It's easy enough to adapt to a character id column with a unique(df1$id) variable.
Here is an option too:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
slice(4:nrow(.))
This gives:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
Alternative:
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate_all(funs(replace(.,is.na(.),0))) %>%
unique()
UPDATE based on other requirements:
Some users suggested to test on this dataframe. Of course this answer assumes you'll look at everything by hand. Might be less useful if you have to look at everything by "hand" but here goes:
df1 <- data.frame(id = rep(c("a", "b","c"), each = 4), val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
df1 %>%
mutate_if(is.factor,as.character) %>%
mutate(val=ifelse(id=="a",0,val)) %>%
slice(4:nrow(.))
This yields:
id val
1 a 0
2 b 1
3 b 2
4 b 2
5 b 3
6 c NA
7 c 2
8 c NA
9 c 3
Here is a base R solution.
res <- lapply(split(df1, df1$id), function(DF){
if(anyNA(DF$val)) {
i <- is.na(DF$val)
DF$val[i] <- 0
DF <- rbind(DF[i & !duplicated(DF[i, ]), ], DF[!i, ])
}
DF
})
res <- do.call(rbind, res)
row.names(res) <- NULL
res
# id val
#1 a 0
#2 b 1
#3 b 2
#4 b 2
#5 b 3
Edit.
A dplyr solution could be the following.
It was tested with the original dataset posted by the OP, with the dataset in Vivek Kalyanarangan's answer and with the dataset in markus' comment, renamed df2 and df3, respectively.
library(dplyr)
na2zero <- function(DF){
DF %>%
group_by(id) %>%
mutate(val = ifelse(is.na(val), 0, val),
crit = val == 0 & duplicated(val)) %>%
filter(!crit) %>%
select(-crit)
}
na2zero(df1)
na2zero(df2)
na2zero(df3)
One may try this :
df1 = data.frame(id = rep(c("a", "b","c"), each = 4),
val = c(NA, NA, NA, NA, 1, 2, 2, 3,NA,2,NA,3))
df1
# id val
#1 a NA
#2 a NA
#3 a NA
#4 a NA
#5 b 1
#6 b 2
#7 b 2
#8 b 3
#9 c NA
#10 c 2
#11 c NA
#12 c 3
Task is to remove all rows corresponding to any id IFF val for the corresponding id is all NAs and add new row with this id and val = 0.
In this example, id = a.
Note : val for c also has NAs but all the val corresponding to c are not NA therefore we need to remove the corresponding row for c where val = NA.
So lets create another column say, val2 which indicates 0 means its all NAs and 1 otherwise.
library(dplyr)
df1 = df1 %>%
group_by(id) %>%
mutate(val2 = if_else(condition = all(is.na(val)),true = 0, false = 1))
df1
# A tibble: 12 x 3
# Groups: id [3]
# id val val2
# <fct> <dbl> <dbl>
#1 a NA 0
#2 a NA 0
#3 a NA 0
#4 a NA 0
#5 b 1 1
#6 b 2 1
#7 b 2 1
#8 b 3 1
#9 c NA 1
#10 c 2 1
#11 c NA 1
#12 c 3 1
Get the list of ids with corresponding val = NA for all.
all_na = unique(df1$id[df1$val2 == 0])
Then remove theids from the dataframe df1 with val = NA.
df1 = na.omit(df1)
df1
# A tibble: 6 x 3
# Groups: id [2]
# id val val2
# <fct> <dbl> <dbl>
# 1 b 1 1
# 2 b 2 1
# 3 b 2 1
# 4 b 3 1
# 5 c 2 1
# 6 c 3 1
And create a new dataframe with ids in all_na and val = 0
all_na_df = data.frame(id = all_na, val = 0)
all_na_df
# id val
# 1 a 0
then combine these two dataframes.
df1 = bind_rows(all_na_df, df1[,c('id', 'val')])
df1
# id val
# 1 a 0
# 2 b 1
# 3 b 2
# 4 b 2
# 5 b 3
# 6 c 2
# 7 c 3
Hope this helps and Edits are most welcomed :-)
I need to fill $Year with missing values of the sequence by the factor of $Country. The $Count column can just be padded out with 0's.
Country Year Count
A 1 1
A 2 1
A 4 2
B 1 1
B 3 1
So I end up with
Country Year Count
A 1 1
A 2 1
A 3 0
A 4 2
B 1 1
B 2 0
B 3 1
Hope that's clear guys, thanks in advance!
This is a dplyr/tidyr solution using complete and full_seq:
library(dplyr)
library(tidyr)
df %>% group_by(Country) %>% complete(Year=full_seq(Year,1),fill=list(Count=0))
Country Year Count
<chr> <dbl> <dbl>
1 A 1 1
2 A 2 1
3 A 3 0
4 A 4 2
5 B 1 1
6 B 2 0
7 B 3 1
library(data.table)
# d is your original data.frame
setDT(d)
foo <- d[, .(Year = min(Year):max(Year)), Country]
res <- merge(d, foo, all.y = TRUE)[is.na(Count), Count := 0]
Similar to #PoGibas' answer:
library(data.table)
# set default values
def = list(Count = 0L)
# create table with all levels
fullDT = setkey(DT[, .(Year = seq(min(Year), max(Year))), by=Country])
# initialize to defaults
fullDT[, names(def) := def ]
# overwrite from data
fullDT[DT, names(def) := mget(sprintf("i.%s", names(def))) ]
which gives
Country Year Count
1: A 1 1
2: A 2 1
3: A 3 0
4: A 4 2
5: B 1 1
6: B 2 0
7: B 3 1
This generalizes to having more columns (besides Count). I guess similar functionality exists in the "tidyverse", with a name like "expand" or "complete".
Another base R idea can be to split on Country, use setdiff to find the missing values from the seq(max(Year)), and rbind them to original data frame. Use do.call to rbind the list back to a data frame, i.e.
d1 <- do.call(rbind, c(lapply(split(df, df$Country), function(i){
x <- rbind(i, data.frame(Country = i$Country[1],
Year = setdiff(seq(max(i$Year)), i$Year),
Count = 0));
x[with(x, order(Year)),]}), make.row.names = FALSE))
which gives,
Country Year Count
1 A 1 1
2 A 2 1
3 A 3 0
4 A 4 2
5 B 1 1
6 B 2 0
7 B 3 1
> setkey(DT,Country,Year)
> DT[setkey(DT[, .(min(Year):max(Year)), by = Country], Country, V1)]
Country Year Count
1: A 1 1
2: A 2 1
3: A 3 NA
4: A 4 2
5: B 1 1
6: B 2 NA
7: B 3 1
Another dplyr and tidyr solution.
library(dplyr)
library(tidyr)
dt2 <- dt %>%
group_by(Country) %>%
do(data_frame(Country = unique(.$Country),
Year = full_seq(.$Year, 1))) %>%
full_join(dt, by = c("Country", "Year")) %>%
replace_na(list(Count = 0))
Here is an approach in base R that uses tapply, do.call, range, and seq, to calculate year sequences. Then constructs a data.frame from the named list that is returned, merges this onto the original which adds the desired rows, and finally fills in missing values.
# get named list with year sequences
temp <- tapply(dat$Year, dat$Country, function(x) do.call(seq, as.list(range(x))))
# construct data.frame
mydf <- data.frame(Year=unlist(temp), Country=rep(names(temp), lengths(temp)))
# merge onto original
mydf <- merge(dat, mydf, all=TRUE)
# fill in missing values
mydf[is.na(mydf)] <- 0
This returns
mydf
Country Year Count
1 A 1 1
2 A 2 1
3 A 3 0
4 A 4 2
5 B 1 1
6 B 2 0
7 B 3 1