I am attempting to have R read across columns by row and evaluate whether values from two adjacent cells are equal. If the values are equal, I want R to count this occurence in a new variable. Here is example data (df):
Var1
Var2
Var3
2
3
3
3
3
3
1
2
3
3
2
1
...and I want to get here:
Var1
Var2
Var3
NewVar
2
3
3
1
3
3
3
2
1
2
3
0
3
2
1
0
One example set of code I have tried out is the following:
df$NewVar <- 0
for (i in 1:2){
if (df[i]==df[i+1]){
df$NewVar <- df$NewVar + 1
}
else{
df$NewVar <- df$NewVar
}
}
This particular set of code just returns 0s in the NewVar variable.
Any sort of help would be much appreciated!
Here's a vectorized solution using rowSums :
df$NewVar <- rowSums(df[-1] == df[-ncol(df)])
df
# Var1 Var2 Var3 NewVar
#1 2 3 3 1
#2 3 3 3 2
#3 1 2 3 0
#4 3 2 1 0
data
df <- structure(list(Var1 = c(2L, 3L, 1L, 3L), Var2 = c(3L, 3L, 2L,
2L), Var3 = c(3L, 3L, 3L, 1L)), class = "data.frame", row.names = c(NA,-4L))
We can use Reduce
df$NewVar <- Reduce(`+`, Map(`==`, df[-1], df[-ncol(df)]))
data
df <- structure(list(Var1 = c(2L, 3L, 1L, 3L), Var2 = c(3L, 3L, 2L,
2L), Var3 = c(3L, 3L, 3L, 1L)), class = "data.frame", row.names = c(NA,-4L))
Related
Despite using R and dplyr on a regular basis, I encountered the issue of not being able to calculate the sum of the absolute differences between all columns:
sum_diff=ABS(A-B)+ABS(B-C)+ABS(C-D)...
A
B
C
D
sum_diff
1
2
3
4
3
2
1
3
4
4
1
2
1
1
2
4
1
2
1
5
I know I could iterate using a for loop over all columns, but given the size of my data frame, I prefer a more elegant and fast solution.
Any help?
Thank you
We may remove the first and last columns, get the difference, and use rowSums on the absolute values in base R. This could be very efficient compared to a package solution
df1$sum_diff <- rowSums(abs(df1[-ncol(df1)] - df1[-1]))
-output
> df1
A B C D sum_diff
1 1 2 3 4 3
2 2 1 3 4 4
3 1 2 1 1 2
4 4 1 2 1 5
Or another option is rowDiffs from matrixStats
library(matrixStats)
rowSums(abs(rowDiffs(as.matrix(df1))))
[1] 3 4 2 5
data
df1 <- structure(list(A = c(1L, 2L, 1L, 4L), B = c(2L, 1L, 2L, 1L),
C = c(3L, 3L, 1L, 2L), D = c(4L, 4L, 1L, 1L)), row.names = c(NA,
-4L), class = "data.frame")
Daata from akrun (many thanks)!
This is complicated the idea is to generate a list of the combinations, I tried it with combn but then I get all possible combinations. So I created by hand.
With this combinations we then could use purrrs map_dfc and do some data wrangling after that:
library(tidyverse)
combinations <-list(c("A", "B"), c("B", "C"), c("C","D"))
purrr::map_dfc(combinations, ~{df <- tibble(a=data[[.[[1]]]]-data[[.[[2]]]])
names(df) <- paste0(.[[1]],"_v_",.[[2]])
df}) %>%
transmute(sum_diff = rowSums(abs(.))) %>%
bind_cols(data)
sum_diff A B C D
<dbl> <int> <int> <int> <int>
1 3 1 2 3 4
2 4 2 1 3 4
3 2 1 2 1 1
4 5 4 1 2 1
data:
data <- structure(list(A = c(1L, 2L, 1L, 4L), B = c(2L, 1L, 2L, 1L),
C = c(3L, 3L, 1L, 2L), D = c(4L, 4L, 1L, 1L)), row.names = c(NA,
-4L), class = "data.frame")
Here is a dplyrs version of #akrun's elegant aproach that calculates the diff of the dataframe with it's shifted variant:
df %>%
mutate(sum_diff = rowSums(abs(identity(.) %>% select(1:last_col(1))
- identity(.) %>% select(2:last_col()))))
And here we have the rowwise variant, which basicly follows the same idea but this time every row is used as a vector that get's substracted by it's shifted self.
df %>%
rowwise() %>%
mutate(sum_diff = map2_int(c_across(1:last_col(1)),
c_across(2:last_col()),
~ abs(.x - .y)) %>% sum())
I would like to find a way to do very similar to this question.
Increment by 1 for every change in column
But i want to restart the counter when var1 = c
using
df$var2 <- with(rle(as.character(df$var1)), rep(seq_along(values), lengths))*
results in column var 2
var1 var2 Should be
a 1 1
a 1 1
1 2 2
0 3 3
b 4 4
b 4 4
b 4 4
c 5 1
1 6 2
1 6 2
In data.table you can use rleid to get a run-length-id for var1 within each group.
library(data.table)
setDT(df)
df[, var2 := rleid(var1), by = cumsum(var1 == "c")]
df
# var1 var2
# 1: a 1
# 2: a 1
# 3: 1 2
# 4: 0 3
# 5: b 4
# 6: b 4
# 7: b 4
# 8: c 1
# 9: 1 2
#10: 1 2
and using dplyr
library(dplyr)
df %>%
group_by(group = cumsum(var1 == "c")) %>%
mutate(var2 = cumsum(var1 != lag(var1, default = first(var1))) + 1)
data
df <- structure(list(var1 = structure(c(3L, 3L, 2L, 1L, 4L, 4L, 4L,
5L, 2L, 2L), .Label = c("0", "1", "a", "b", "c"), class = "factor")),
class = "data.frame", row.names = c(NA, -10L))
We can use the OP's code with rle in base R with ave
df$var2 <- with(df, as.integer(ave(as.character(var1), cumsum(var1 == 'c'),
FUN = function(x) with(rle(x), rep(seq_along(values), lengths)))))
df$var2
#[1] 1 1 2 3 4 4 4 1 2 2
data
df <- structure(list(var1 = structure(c(3L, 3L, 2L, 1L, 4L, 4L, 4L,
5L, 2L, 2L), .Label = c("0", "1", "a", "b", "c"), class = "factor")),
class = "data.frame", row.names = c(NA,
-10L))
Hello I have the data frame and I need to remove all the rows with max values from each columns.
Example
A B C
1 2 3 5
2 4 1 1
3 1 4 3
4 2 1 1
So the output is:
A B C
4 2 1 1
Is there any quick way to do this?
We can do this with %in%
df1[!seq_len(nrow(df1)) %in% sapply(df1, which.max),]
# A B C
#4 2 1 1
If there are ties for maximum values in each row, then do
df1[!Reduce(`|`, lapply(df1, function(x) x== max(x))),]
df[-sapply(df, which.max),]
# A B C
#4 2 1 1
DATA
df = structure(list(A = c(2L, 4L, 1L, 2L), B = c(3L, 1L, 4L, 1L),
C = c(5L, 1L, 3L, 1L)), .Names = c("A", "B", "C"),
class = "data.frame", row.names = c(NA,-4L))
I have data that looks like this
ID v1 v2
1 1 0
2 0 1
3 1 0
3 0 1
4 0 1
I want to replace all values with 'NA' if the ID occurs more than once in the dataframe. The final product should look like this
ID v1 v2
1 1 0
2 0 1
3 NA NA
3 NA NA
4 0 1
I could do this by hand, but I want R to detect all the duplicate cases (in this case two times ID '3') and replace the values with 'NA'.
Thanks for your help!
You could use duplicated() from either end, and then replace.
idx <- duplicated(df$ID) | duplicated(df$ID, fromLast = TRUE)
df[idx, -1] <- NA
which gives
ID v1 v2
1 1 1 0
2 2 0 1
3 3 NA NA
4 3 NA NA
5 4 0 1
This will also work if the duplicated IDs are not next to each other.
Data:
df <- structure(list(ID = c(1L, 2L, 3L, 3L, 4L), v1 = c(1L, 0L, 1L,
0L, 0L), v2 = c(0L, 1L, 0L, 1L, 1L)), .Names = c("ID", "v1",
"v2"), class = "data.frame", row.names = c(NA, -5L))
One more option:
df1[df1$ID %in% df1$ID[duplicated(df1$ID)], -1] <- NA
#> df1
# ID v1 v2
#1 1 1 0
#2 2 0 1
#3 3 NA NA
#4 3 NA NA
#5 4 0 1
data
df1 <- structure(list(ID = c(1L, 2L, 3L, 3L, 4L), v1 = c(1L, 0L, 1L,
0L, 0L), v2 = c(0L, 1L, 0L, 1L, 1L)), .Names = c("ID", "v1",
"v2"), class = "data.frame", row.names = c(NA, -5L))
Here is a base R method
# get list of repeated IDs
repeats <- rle(df$ID)$values[rle(df$ID)$lengths > 1]
# set the corresponding variables to NA
df[, -1] <- sapply(df[, -1], function(i) {i[df$ID %in% repeats] <- NA; i})
In the first line, we use rle to extract repeated IDs. In the second, we use sapply to loop through non-ID variables and replace IDs that repeat with NA for each variable.
Note that this assumes that the data set is sorted by ID. This may be accomplished with the order function. (df <- df[order(df$ID),]).
If the dataset is very large, you might break up the first function into two steps to avoid computing the rle twice:
dfRle <- rle(df$ID)
repeats <- dfRle$values[dfRle$lengths > 1]
data
df <- read.table(header=T, text="ID v1 v2
1 1 0
2 0 1
3 1 0
3 0 1
4 0 1")
I have a dataframe which looks like -
Id Result
A 1
B 2
C 1
B 1
C 1
A 2
B 1
B 2
C 1
A 1
B 2
Now I need to calculate how many 1's and 2's are there for each Id and then select the number whose frequency of occurrence is the greatest.
Id Result
A 1
B 2
C 1
How can I do that? I have tried using the table function in some way but not able to use it effectively. Any help would be appreciated.
Here you can use aggregate in one step:
df <- structure(list(Id = structure(c(1L, 2L, 3L, 2L, 3L, 1L, 2L, 2L,
3L, 1L, 2L), .Label = c("A", "B", "C"), class = "factor"),
Result = c(1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 2L)),
.Names = c("Id", "Result"), class = "data.frame", row.names = c(NA, -11L)
)
res <- aggregate(Result ~ Id, df, FUN=function(x){which.max(c(sum(x==1), sum(x==2)))})
res
Result:
Id Result
1 A 1
2 B 2
3 C 1
With data.table you can try (df is your data.frame):
require(data.table)
dt<-as.data.table(df)
dt[,list(times=.N),by=list(Id,Result)][,list(Result=Result[which.max(times)]),by=Id]
# Id Result
#1: A 1
#2: B 2
#3: C 1
Using dplyr, you can try
library(dplyr)
df %>% group_by(Id, Result) %>% summarize(n = n()) %>% group_by(Id) %>%
filter(n == max(n)) %>% summarize(Result = Result)
Id Result
1 A 1
2 B 2
3 C 1
An option using table and ave
subset(as.data.frame(table(df1)),ave(Freq, Id, FUN=max)==Freq, select=-3)
# Id Result
# 1 A 1
# 3 C 1
# 5 B 2