I want to remove columns that have only a unique value.
First, I try it for a single column and it works:
data %/%
select_if(length(unique(data$policy_id)) > 1)
then I try it for multiple columns as below:
data %/%
select_if(length(unique(data[, c("policy_date", "policy_id"])) > 1)
but it does not work. I think it is a conceptual mistake due to my lack of experience.
thanks in advance
You can use select(where()).
Suppose I have a data frame like this:
df <- data.frame(A = LETTERS[1:5], B = 1:5, C = 2)
df
#> A B C
#> 1 A 1 2
#> 2 B 2 2
#> 3 C 3 2
#> 4 D 4 2
#> 5 E 5 2
Then I can do:
df %>% select(where(~ n_distinct(.) > 1))
#> A B
#> 1 A 1
#> 2 B 2
#> 3 C 3
#> 4 D 4
#> 5 E 5
Some base R options:
Using lengths + unique + sapply
subset(df,select = lengths(sapply(df,unique))>1)
Using Filter + length + unique
Filter(function(x) length(unique(x))>1,df)
Does this work:
> df <- data.frame(col1 = 1:10,
+ col2 = rep(10,10),
+ col3 = round(rnorm(10,1)))
> df
col1 col2 col3
1 1 10 1
2 2 10 0
3 3 10 1
4 4 10 1
5 5 10 1
6 6 10 0
7 7 10 2
8 8 10 1
9 9 10 1
10 10 10 1
> df %>% select_if(~length(unique(.)) > 1)
col1 col3
1 1 1
2 2 0
3 3 1
4 4 1
5 5 1
6 6 0
7 7 2
8 8 1
9 9 1
10 10 1
>
Another option would be to use purrr:
df %>% purrr::keep(~all(n_distinct(.) > 1))
df %>% purrr::keep(~all(length(unique(.)) > 1))
df %>% purrr::discard(~!all(n_distinct(.) > 1))
df %>% purrr::discard(~!all(length(unique(.)) > 1))
Mixing table with apply generates the same output.
df[, apply(df, 2, function(i) length(table(i)) > 1)]
df <- data.frame(A = LETTERS[1:5], B = 1:5, C = 2)
An option with base R
df[sapply(df, function(x) length(unique(x))) > 1]
data
df <- data.frame(A = LETTERS[1:5], B = 1:5, C = 2)
I want to fill one step up with one variable in a data frame
> id <- rep(1:3,each=2)
> trt <- rep(c("A","B"),3)
> score <- c("1", "","", 3, "",6)
> df <- data.frame(id,trt,score)
> df
id trt score
1 1 A 1
2 1 B
3 2 A
4 2 B 3
5 3 A
6 3 B 6
>
I want it to look like this:
> id <- rep(1:3,each=2)
> trt <- rep(c("A","B"),3)
> score <- c(1, "",3, 3, 6,6)
> df <- data.frame(id,trt,score)
> df
id trt score
1 1 A 1
2 1 B
3 2 A 3
4 2 B 3
5 3 A 6
6 3 B 6
I know this code fill in the columns up, but I just want it to fill one step up, is that possible to do?
library(tidyr)
> library(dplyr)
> df %>% fill(score, .direction="up")
To use fill we need NA's while you have empty string values. We can conditional replace blank values with NA for only one row above non-blank values and then use fill
library(dplyr)
df %>%
mutate(score = replace(score, which(score != "") - 1, NA)) %>%
tidyr::fill(score, .direction = "up")
# id trt score
#1 1 A 1
#2 1 B
#3 2 A 3
#4 2 B 3
#5 3 A 6
#6 3 B 6
An alternative and simple base R option would be
inds <- which(df$score != '')
inds <- inds[inds > 1]
df$score[inds - 1] <- df$score[inds]
I've a dataframe which I want to add a row on the basis of the following conditions. The conditions are when column a is equal to C and column b is equal to 3 or 5.
Here is my dataframe
df <- data.frame(a = c("A", "B", "C", "D", "C", "A", "C", "E"),
b = c(seq(8)), stringsAsFactors = TRUE)
Whenever the condition is TRUE I want to add a row below where the condition is met add 3. I have tried the following
rbind(df, data.frame(a="add", b = "3"))
# a b
# 1 A 1
# 2 B 2
# 3 C 3
# 4 D 4
# 5 C 5
# 6 A 6
# 7 C 7
# 8 E 8
# 9 add 3
This is not the output I want. The output I want is
# a b
# 1 A 1
# 2 B 2
# 3 C 3
# 4 add 3
# 5 D 4
# 6 C 5
# 7 add 3
# 8 A 6
# 9 C 7
# 10 E 8
How can I do that? I am new to R and thank you for your help.
lens = ifelse(df$b %in% c(3, 5) & df$a == "C", 2, 1)
ind = rep(1:NROW(df), lens)
df2 = df[ind,]
df2$a = as.character(df2$a)
df2$a[cumsum(lens)[which(lens == 2)]] = "add"
df2$b[cumsum(lens)[which(lens == 2)]] = 3
df2
# a b
#1 A 1
#2 B 2
#3 C 3
#3.1 add 3
#4 D 4
#5 C 5
#5.1 add 3
#6 A 6
#7 C 7
#8 E 8
A solution using the tidyverse package.
library(tidyverse)
df2 <- df %>%
mutate(Group = lag(cumsum(a == "C" & b %in% c(3, 5)), default = FALSE)) %>%
group_split(Group) %>%
map_dfr(~ .x %>% bind_rows(tibble(a = "add", b = 3))) %>%
slice(-n()) %>%
select(-Group)
df2
# # A tibble: 10 x 2
# a b
# <chr> <dbl>
# 1 A 1
# 2 B 2
# 3 C 3
# 4 add 3
# 5 D 4
# 6 C 5
# 7 add 3
# 8 A 6
# 9 C 7
# 10 E 8
In base R, we can find out position where a = "c" and b is 3 or 5. Repeat those rows in the dataframe and replace them with required values.
pos <- which(df$a == "C" & df$b %in% c(3, 5))
df <- df[sort(c(seq(nrow(df)), pos)), ]
df[seq_along(pos) + pos, ] <- list("add", 3)
row.names(df) <- NULL
df
# a b
#1 A 1
#2 B 2
#3 C 3
#4 add 3
#5 D 4
#6 C 5
#7 add 3
#8 A 6
#9 C 7
#10 E 8
data
df <- data.frame(a = c("A", "B", "C", "D", "C", "A", "C", "E"),
b = c(seq(8)), stringsAsFactors = FALSE)
I am trying to fill in blank cells with the value of rows above. Similar to na.locf function, but I have a pattern that needs to be matched. I don't necessarily know how many rows between new values (i.e betweem a,b and c,d).
I have used the na.locf and searched around for a solution to no avail.
df <- df <- data.frame(col1 = c("a","b", NA, NA, NA, NA, "c", "d", NA, NA))
df
# col1
# 1 a
# 2 b
# 3 <NA>
# 4 <NA>
# 5 <NA>
# 6 <NA>
# 7 c
# 8 d
# 9 <NA>
# 10 <NA>
Solution I would like:
df
col1
a
b
a
b
a
b
c
d
c
d
ave(df$col1,
with(rle(!is.na(df$col1)), rep(cumsum(values), lengths)),
FUN = function(x){
rep(x[!is.na(x)], length.out = length(x))
})
# [1] a b a b a b c d c d
Here's way with dplyr. You can drop the group column if needed. -
df %>%
group_by(group = cumsum(is.na(lag(col1)) & !is.na(col1))) %>%
mutate(
col1 = rep(col1[!is.na(col1)], length.out = n())
) %>%
ungroup()
# A tibble: 10 x 2
col1 group
<chr> <int>
1 a 1
2 b 1
3 a 1
4 b 1
5 a 1
6 b 1
7 c 2
8 d 2
9 c 2
10 d 2
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 :-)