Append dataFrame columns to other columns with different names and order? - r

Question

I am struggling with reordering a dataFrame in R.
My dataFrame has data coming from two different sensors. So in the beginning every column has a name with the syntax "sensor number.sample number". The rowname is a coordinate of each sample.
Sadly the columns are not ordered with an ascending sample number.
How can I make an automatic ordering where after number 1 comes 2 and not 10?
With correct ordered columns I would like to cut all columns of the second sensor and append it under the rows from the first sensor. This is also tricky as the number of columns of each sensor varies in the reality.
To distinguish between both sensors I would add a postfix "a" or "b" for the new rownames.
Here my problem is that I know "rbind" but it requires identical column names, I cannot provide here. And I would also need to select the columns manually as I have no clue how to automatically select all of the second sensor.
My idea for the moment is to make subsets for each sensor, rename the columns and then use rbind with both subsets. Is this a good idea?
The rownames I then could modify with paste().
I now present simplified frames as the original is quite big. So the numbers (c(1:3)) are just exemplary.
This is how my dataFrame looks at the beginning:
myDf = data.frame(a.10= c(1:3),a.11= c(1:3),a.12= c(1:3),a.13= c(1:3),a.2= c(1:3),a.3= c(1:3),a.4= c(1:3),a.5= c(1:3),a.6= c(1:3),a.7= c(1:3),a.8= c(1:3),a.9= c(1:3),
b.1= c(1:3),b.10= c(1:3),b.11= c(1:3),b.2= c(1:3),b.3= c(1:3),b.4= c(1:3),b.5= c(1:3),b.6= c(1:3),b.7= c(1:3),b.8= c(1:3),b.9= c(1:3))
My goal is to transform the dataFrame that is looks like that:
desiredDf =data.frame(n9=rep(c(1:3),2), n10=rep(c(1:3),2), n11=rep(c(1:3),2), n12=c(c(1:3),NA, NA, NA), n13=c(c(1:3), NA, NA, NA))
rownames(desiredDf)<-(c("1a","2a","3a","1b","2b","3b"))
Thank you very much!

Answer 1

Here is an option.
library(tidyverse)
myDF2 <- myDf %>% gather(measure, result, a.10:b.9) %>%
separate(measure, into = c("letter", "number"), sep = "\\.") %>%
group_by(letter, number)%>%
mutate(n = row_number()) %>%
unite(col, n, letter, sep = "") %>%
ungroup() %>%
arrange(as.numeric(number))%>%
mutate(number = paste0("n", number))%>%
mutate(number = factor(number, levels = unique(number)))%>%
spread(number, result)%>%
arrange(col)
row.names(myDF2) <- myDF2$col
myDF2$col <- NULL

Answer 2

Convert the row names to a column, reshape into long form and separate the key, i.e. the original column names, into columns group and no converting the latter to numeric. Sort, reshape back to wide form, sort again, combine the rowname and group and preface each column name with n.
library(dplyr)
library(tibble)
library(tidyr)
myDf %>%
rownames_to_column %>%
gather(key, value, -rowname) %>%
separate(key, c("group", "no"), convert = TRUE) %>%
arrange(group, no) %>%
spread(no, value) %>%
arrange(group, rowname) %>%
unite(rowname, rowname, group, sep = "") %>%
column_to_rownames %>%
rename_all(~ paste0("n", .))
giving:
n1 n2 n3 n4 n5 n6 n7 n8 n9 n10 n11 n12 n13
1a NA 1 1 1 1 1 1 1 1 1 1 1 1
2a NA 2 2 2 2 2 2 2 2 2 2 2 2
3a NA 3 3 3 3 3 3 3 3 3 3 3 3
1b 1 1 1 1 1 1 1 1 1 1 1 NA NA
2b 2 2 2 2 2 2 2 2 2 2 2 NA NA
3b 3 3 3 3 3 3 3 3 3 3 3 NA NA
Note
Above we used this for myDf, the input.
myDf <-
structure(list(a.10 = 1:3, a.11 = 1:3, a.12 = 1:3, a.13 = 1:3,
a.2 = 1:3, a.3 = 1:3, a.4 = 1:3, a.5 = 1:3, a.6 = 1:3, a.7 = 1:3,
a.8 = 1:3, a.9 = 1:3, b.1 = 1:3, b.10 = 1:3, b.11 = 1:3,
b.2 = 1:3, b.3 = 1:3, b.4 = 1:3, b.5 = 1:3, b.6 = 1:3, b.7 = 1:3,
b.8 = 1:3, b.9 = 1:3), class = "data.frame", row.names = c(NA,
-3L))