Based on my previous question, I need help with using the mapply function correctly.
x <- data.frame(a = seq(1,3), b = seq(2,4), c = seq(3,5), d = seq(4,6), b2 = seq(5,7), c2 = seq(6,8), d2 = seq(7,9))
# a b c d b2 c2 d2
# 1 2 3 4 5 6 7
# 2 3 4 5 6 7 8
# 3 4 5 6 7 8 9
My goal is to look at the columns b2 to d2 and, based on their values, change the values in columns b to d respectively. I can do this for a single column quite easily:
x[which(x$b2 == 7),][b] <- NA_real_
My problem is that I want this applied across all my columns but I don't know how to convert this single column formula to work on multiple columns. I tried:
onez <- c(2:4)
twoz <- c(5:7)
f <- function(df, ones, twos) {
df[which(df[,twos] == 7),][ones] <- NA_real_
}
mapply(f, df = x, ones = onez, twos = twoz)
But I'm getting error messages (incorrect dimensions etc) and I see that my function is messy but I lack the knowledge how to fix it.
One way to do it is to tell it to:
Get the subset of the data frame with columns 5, 6, 7: x[5:7]
Check from that subset which values satisfy your condition: x[5:7] == 7
Replace those values with NA: ... <- NA
This gives the following,
x[5:7][x[5:7] == 7] <- NA
x
# a b c d b2 c2 d2
#1 1 2 3 4 5 6 NA
#2 2 3 4 5 6 NA 8
#3 3 4 5 6 NA 8 9
If you want the NAs to be replaced at x[2:4], then you can do,
x[2:4][x[5:7] == 7] <- NA
x
# a b c d b2 c2 d2
#1 1 2 3 NA 5 6 7
#2 2 3 NA 5 6 7 8
#3 3 NA 5 6 7 8 9
Related
lets say data is 'ab':
a <- c(1,2,3,NA,5,NA)
b <- c(5,NA,4,NA,NA,6)
ab <-c(a,b)
I would like to have new variable which is sum of the two but keeping NA's as follows:
desired output:
ab$c <-(6,2,7,NA,5,6)
so addition of number + NA should equal number
I tried following but does not work as desired:
ab$c <- a+b
gives me : 6 NA 7 NA NA NA
Also don't know how to include "na.rm=TRUE", something I was trying.
I would also like to create third variable as categorical based on cutoff <=4 then event 1, otherwise 0:
desired output:
ab$d <-(1,1,1,NA,0,0)
I tried:
ab$d =ifelse(ab$a<=4|ab$b<=4,1,0)
print(ab$d)
gives me logical(0)
Thanks!
a <- c(1,2,3,NA,5,NA)
b <- c(5,NA,4,NA,NA,6)
dfd <- data.frame(a,b)
dfd$c <- rowSums(dfd, na.rm = TRUE)
dfd$c <- ifelse(is.na(dfd$a) & is.na(dfd$b), NA_integer_, dfd$c)
dfd$d <- ifelse(dfd$c >= 4, 1, 0)
dfd
a b c d
1 1 5 6 1
2 2 NA 2 0
3 3 4 7 1
4 NA NA NA NA
5 5 NA 5 1
6 NA 6 6 1
In my code below, I was wondering if there might be a way to add the z1 vector to data.frame d1 such that we can achieve my Desired_Output using Base R or tidyverse?
This is a toy example. Thus, d1 can have any number of rows and columns and z1 vector can have any number elements. Thus, a functional answer applicable to other data.frames is highly appreciated.
d1 <- data.frame(b = 1:5, SE = 2:6)
z1 <- c(2.3, 5.4)
d1$tau <- ""
Desired_Output =
" b SE tau
1 2
2 3
3 4
4 5
5 6
2.3
5.4"
You may use dplyr::bind_rows or data.table::rbindlist
d1 <- data.frame(b = 1:5, SE = 2:6)
z1 <- c(2.3, 5.4)
d2 <- data.frame(tau = z1)
dplyr::bind_rows(d1, d2)
# b SE tau
#1 1 2 NA
#2 2 3 NA
#3 3 4 NA
#4 4 5 NA
#5 5 6 NA
#6 NA NA 2.3
#7 NA NA 5.4
With data.table -
data.table::rbindlist(list(d1, d2), fill = TRUE)
The d1 data frame has 5 rows and two columns. To add a column, it, too, must have 5 rows. However, because it is required for the z vector to occupy rows 6 and 7, those rows must first be added to d1:
d1 <- data.frame(b = 1:5, SE = 2:6)
d1[6:7,] <- NA
d1$tau <- c(rep(NA,5),2.3,5.4)
d1
#> b SE tau
#> 1 1 2 NA
#> 2 2 3 NA
#> 3 3 4 NA
#> 4 4 5 NA
#> 5 5 6 NA
#> 6 NA NA 2.3
#> 7 NA NA 5.4
How can I insert blank rows every 3 existing rows in a data frame?
After a web scraping process I get a dataframe with the information I need, however the final excel format requires that I add a blank row every 3 rows. I have searched the web for help but have not found a solution yet.
With hypothetical data, the structure of my data frame is as follows:
mi_df <- data.frame(
"ID" = rep(1:3,c(3,3,3)),
"X" = as.character(c("a", "a", "a", "b", "b", "b", "c", "c", "c")),
"Y" = seq(1,18, by=2)
)
mi_df
ID X Y
1 1 a 1
2 1 a 3
3 1 a 5
4 2 b 7
5 2 b 9
6 2 b 11
7 3 c 13
8 3 c 15
9 3 c 17
The result I hope for is something like this
ID X Y
1 1 a 1
2 1 a 3
3 1 a 5
4
5 2 b 7
6 2 b 9
7 2 b 11
8
9 3 c 13
10 3 c 15
11 3 c 17
If the indices of a data frame contain NA, then the output will have NA rows. So my goal is to create a vector like 1 2 3 NA 4 5 6 NA ... and set it as the indices of mi_df.
cut <- rep(1:(nrow(mi_df)/3), each = 3)
mi_df[sapply(split(1:nrow(mi_df), cut), c, NA), ]
# ID X Y
# 1 1 a 1
# 2 1 a 3
# 3 1 a 5
# NA NA <NA> NA
# 4 2 b 7
# 5 2 b 9
# 6 2 b 11
# NA.1 NA <NA> NA
# 7 3 c 13
# 8 3 c 15
# 9 3 c 17
# NA.2 NA <NA> NA
If nrow(mi_df) is not a multiple of 3, then the following is a general solution:
# Version 1
cut <- rep(1:ceiling(nrow(mi_df)/3), each = 3, len = nrow(mi_df))
mi_df[Reduce(c, lapply(split(1:nrow(mi_df), cut), c, NA)), ]
# Version 2
cut <- rep(1:ceiling(nrow(mi_df)/3), each = 3, len = nrow(mi_df))
mi_df[Reduce(function(x, y) c(x, NA, y), split(1:nrow(mi_df), cut)), ]
Don't mind the NA in the output because some functions which write data to an excel file have an optional argument controls if NA values are converted to strings or be empty. E.g.
library(openxlsx)
write.xlsx(df, "test.xlsx", keepNA = FALSE) # defaults to FALSE
tmp <- split(mi_df, rep(1:(nrow(mi_df) / 3), each = 3))
# or split(mi_df, ggplot2::cut_width(seq_len(nrow(mi_df)), 3, center = 2))
do.call(rbind, lapply(tmp, function(x) { x[4, ] <- NA; x }))
ID X Y
1.1 1 a 1
1.2 1 a 3
1.3 1 a 5
1.4 NA <NA> NA
2.4 2 b 7
2.5 2 b 9
2.6 2 b 11
2.4.1 NA <NA> NA
3.7 3 c 13
3.8 3 c 15
3.9 3 c 17
3.4 NA <NA> NA
You can make empty rows like you show by assigning an empty character vector ("") instead of NA, but this will convert your columns to character, and I wouldn't recommend it.
My recommendation is somewhat different from all the other answers: don't make a mess of your dataset inside R . Use the existing packages to write to designated rows in an Excel workbook. For example, with the package xlConnect, the method writeWorksheet (called from writeWorksheetToFile ) includes these arguments:
object The workbook to write to data Data to write
sheet The name or index of the sheet to write to
startRow Index of the first row to write to. The default is startRow = 1.
startCol Index of the first column to write to. The default is startCol = 1.
So if you simply set up a loop that writes 3 rows of your data file at a time, then moves the row index down by 4 and writes the next 3 rows, etc., you're all set.
Here's one method.
Splits into list by ID, adds empty row, then binds list back into data frame.
mi_df2 <- do.call(rbind,Map(rbind,split(mi_df,mi_df$ID),rep("",3)))
rownames(mi_df2) <- NULL
This is my sample data.
index <- c(1,2,3,4,5,6,7,8,9,10)
a <- c('a','b','c',NA,'D','e',NA,'g','h','i')
data <- data.frame(index,a)
What I would like to is create a new column name where only 'a' and 'b' stay. All others like 'c','d','e'...will be tagged as others, while NA stays as NA.
data$name = ifelse(!grepl('(a|b)',data$a),'others',data$name)
I tried to use the grepl function and it seems it is not working with data with missing values
In base R:
data$res <- as.character(data$a)
data$res[! data$a %in% c("a","b") & !is.na(data$a)] <- "Other"
data
# index a res
# 1 1 a a
# 2 2 b b
# 3 3 c Other
# 4 4 <NA> <NA>
# 5 5 D Other
# 6 6 e Other
# 7 7 <NA> <NA>
# 8 8 g Other
# 9 9 h Other
# 10 10 i Other
Note that the new column is of type character here.
Using dplyr and its recode function, you could do
data %>% mutate(name=recode(a, a="a", b="b", .default="other"))
# index a name
# 1 1 a a
# 2 2 b b
# 3 3 c other
# 4 4 <NA> <NA>
# 5 5 D other
# 6 6 e other
# 7 7 <NA> <NA>
# 8 8 g other
# 9 9 h other
# 10 10 i other
With a more complicated match, you migth use case_when instead
data %>% mutate(name=case_when(
is.na(a) ~ NA_character_,
a %in% c("a","b") ~ as.character(a),
TRUE ~ "other"))
I have a problem figuering this out:
suppose this is how my data looks like:
Num condition y
1 a 1
2 a 2
3 a 3
4 b 4
5 b 5
6 b 6
7 c 7
8 c 8
9 c 9
10 b 10
11 b 11
12 b 12
I now want to make calculation (e.g., mean) on b, depending on whether value was in the row before b, in this example a or c?
Thanks for any help!!!
Angelika
Is this what you want?
# in order to separate between different runs of condition 'b',
# get length and value of runs of equal values of 'condition'
rl <- rle(x = df$condition)
df$run <- rep(x = seq_len(length(rl$lengths)), times = rl$lengths)
# calculate sum of y, on data grouped by condition and run, and where condition is 'b'
aggregate(y ~ condition + run, data = df, subset = condition == "b", sum)
You can add a "lagged" condition column to your dataframe (assuming DF) using
> DF <- within(DF, lag_cond <- c(NA, head(as.character(condition), -1)))
Result:
Num condition y lag_cond
1 a 1 <NA>
2 a 2 a
3 a 3 a
4 b 4 a
5 b 5 b
6 b 6 b
7 c 7 b
8 c 8 c
9 c 9 c
10 b 10 c
11 b 11 b
12 b 12 b
Now you can identify rows you want like this:
> DF[with(DF, condition=="b" & lag_cond %in% c("a","c")),]
Num condition y lag_cond
4 b 4 a
10 b 10 c