R: manipulate dataframe cell using named numeric vector - r

I want to (manually) overwrite cells in column x of the dataframe df. But R produces an error. Consider
m = 1:2
n = 3:4
names(m)= c("o", "we")
names(n)= c("bn","lt")
s = c( "bb", "cc")
b = c( FALSE, TRUE)
df = data.frame( s, b)
df$x= list(m,n)
Now replace column x first row:
k = 5:6
names(k)= c("jh","jh")
df[1,"x"] = k ## error occurs here


Try this: df$x[[1]] = k
Since you are using a data frame you can not speak of columns and rows.. so you have to access the elements of the data frame with $ instead of "x".
Further, you are using lists in your data frame. You should access the elements of a list with [[1]] or any other position of the element in the list.
Hope this helps.

Related

How can I save different tables into one csv on the same tab above each other?

Let's say I have 4 tables, and I want to put them in a csv file like this:
How would I do that? I want two columns of tables and any number of rows, and I want to be able to define where which table is and keep 1 column between the two columns and two rows between each row of tables as depicted.

Put your 4 data frames into a named list L such as the test data shown below based on the built-in BOD data frame.
As in the question, we assume that the sum of the number of columns in the first two data frames equals the sum of the number of columns in the last two data frames.
Define a fmt function which inputs a name from the list L and outputs a character ts series with the name at top. We use ts series because those allow cbind to be used with arguments having different numbers of rows. Finally use write.table to output the result. We just output it to the console but you can modify the write.table statement appropriately to output to a file.
No packages are used.
# test data
b <- BOD; rownames(b) <- letters[1:nrow(BOD)]
L <- list(d1 = b[1:2, ], d2 = b[1:3, ], d3 = b[1:4, ], d4 = b[1:5,])
fmt <- function(nm, DF = L[[nm]]) {
m <- cbind(rownames(DF), sapply(DF, as.character))
as.ts(rbind(c(nm, rep("", ncol(DF))), colnames(m), m))
}
m1 <- cbind(fmt("d1"), "", fmt("d2"))
m2 <- cbind(fmt("d3"), "", fmt("d4"))
m <- rbind(m1, "", m2)
write.table(m, sep = ",", na = "",
row.names = FALSE, col.names = FALSE, quote = FALSE)
giving:
d1,,,,d2,,
,Time,demand,,,Time,demand
a,1,8.3,,a,1,8.3
b,2,10.3,,b,2,10.3
,,,,c,3,19
,,,,,,
d3,,,,d4,,
,Time,demand,,,Time,demand
a,1,8.3,,a,1,8.3
b,2,10.3,,b,2,10.3
c,3,19,,c,3,19
d,4,16,,d,4,16
,,,,e,5,15.6

Create a Column with Unique values from Lists Columns

I have a dataset on Rstudio made of columns that contains lists inside them. Here is an example where column "a" and column "c" contain lists in each row.
¿What I am looking for?
I need to create a new column that collects unique values from columns a b and c and that skips NA or null values
Expected result is column "desired_result".
test <- tibble(a = list(c("x1","x2"), c("x1","x3"),"x3"),
b = c("x1", NA,NA),
c = list(c("x1","x4"),"x4","x2"),
desired_result = list(c("x1","x2","x4"),c("x1","x3","x4"),c("x2","x3")))
What i have tried so far?
I tried the following but do not produces the expected result as in column "desired_result
test$attempt_1_ <-lapply(apply((test[, c("a","b","c"), drop = T]),
MARGIN = 1, FUN= c, use.names= FALSE),unique)

We may use pmap to loop over each of the corresponding elements of 'a' to 'c', remove the NA (na.omit) and get the unique values to store as a list in 'desired_result'
library(dplyr)
library(purrr)
test <- test %>%
mutate(desired_result2 = pmap(across(a:c), ~ sort(unique(na.omit(c(...))))))
-checking with OP's expected
> all.equal(test$desired_result, test$desired_result2)
[1] TRUE

Returning specific values within a row

I have 1 row of data and 50 columns in the row from a csv which I've put into a dataframe. The data is arranged across the spreadsheet like this:
"FSEG-DFGS-THDG", "SGDG-SGRE-JJDF", "DIDC-DFGS-LEMS"...
How would I select only the middle part of each element (eg, "DFGS" in the 1st one, "SGRE" in the second etc), count their occurances and display the results?
I have tried using the strsplit function but I couldn't get it to work for the entire row of data. I'm thinking a loop of some kind might be what I need

You can do unlist(strsplit(x, '-'))[seq(2, length(x)*3, 3)] (assuming your data is consistently of the form A-B-C).
# E.g.
fun <- function(x) unlist(strsplit(x, '-'))[seq(2, length(x)*3, 3)]
fun(c("FSEG-DFGS-THDG", "SGDG-SGRE-JJDF", "DIDC-DFGS-LEMS"))
# [1] "DFGS" "SGRE" "DFGS"
Edit
# Data frame
df <- structure(list(a = "FSEG-DFGS-THDG", b = "SGDG-SGRE-JJDF", c = "DIDC-DFGS-LEMS"),
class = "data.frame", row.names = c(NA, -1L))
fun(t(df[1,]))
# [1] "DFGS" "SGRE" "DFGS"

First we create a function strng() and then we apply() it on every column of df. strsplit() splits a string by "-" and strng() returns the second part.
df = data.frame(a = "ab-bc-ca", b = "gn-bc-ca", c = "kj-ll-mn")
strng = function(x) {
strsplit(x,"-")[[1]][2]
}
# table() outputs frequency of elements in the input
table(apply(df, MARGIN = 2, FUN = strng))
# output: bc ll
2 1

cbind equally named vectors in multiple data.frames in a list to a single data.frame

I have a list similar to this one:
set.seed(1602)
l <- list(data.frame(subst_name = sample(LETTERS[1:10]), perc = runif(10), crop = rep("type1", 10)),
data.frame(subst_name = sample(LETTERS[1:7]), perc = runif(7), crop = rep("type2", 7)),
data.frame(subst_name = sample(LETTERS[1:4]), perc = runif(4), crop = rep("type3", 4)),
NULL,
data.frame(subst_name = sample(LETTERS[1:9]), perc = runif(9), crop = rep("type5", 9)))
Question: How can I extract the subst_name-column of each data.frame and combine them with cbind() (or similar functions) to a new data.frame without messing up the order of each column? Additionally the columns should be named after the corresponding crop type (this is possible 'cause the crop types are unique for each data.frame)
EDIT: The output should look as follows:
Having read the comments I'm aware that within R it doesn't make much sense but for the sake of having alook at the output the data.frame's View option is quite handy.

With the help of this SO-Question I came up with the following sollution. (There's probably room for improvement)
a <- lapply(l, '[[', 1) # extract the first element of the dfs in the list
a <- Filter(function(x) !is.null(unlist(x)), a) # remove NULLs
a <- lapply(a, as.character)
max.length <- max(sapply(a, length))
## Add NA values to list elements
b <- lapply(a, function(v) { c(v, rep(NA, max.length-length(v)))})
e <- as.data.frame(do.call(cbind, d))
names(e) <- unlist(lapply(lapply(lapply(l, '[[', "crop"), '[[', 2), as.character))

It is not really correct to do this with the given example because the number of rows is not the same in each one of the list's data frames . But if you don't care you can do:
nullElements = unlist(sapply(l,is.null))
l = l[!nullElements] #delete useless null elements in list
columns=lapply(l,function(x) return(as.character(x$subst_name)))
newDf = as.data.frame(Reduce(cbind,columns))
If you don't want recycled elements in the columns you can do
for(i in 1:ncol(newDf)){
colLength = nrow(l[[i]])
newDf[(colLength+1):nrow(newDf),i] = NA
}
newDf = newDf[1:max(unlist(sapply(l,nrow))),] #remove possible extra NA rows
Note that I edited my previous code to remove NULL entries from l to simplify things

R: Looping through list of dataframes in a vector

I have a dataset where I only want to loop through certain columns in a dataframe one at a time to create a graph. The structure of my dataframe consists of data that I parsed from a larger dataset into a vector containing multiple dataframes.
I want to call one column from one dataframe in the vector. I want to loop on the dataframes to call each column.
See example below:
d1 <- data.frame(y1=c(1,2,3),y2=c(4,5,6))
d2 <- data.frame(y1=c(3,2,1),y2=c(6,5,4))
my.list <- list(d1, d2)
All I have to work with is my.list
How would I do this?

You can use lapply to plot each of the individual data frames in your list. For example,
d1 <- data.frame(y1=c(1,2,3),y2=c(4,5,6),y3=c(7,8,9))
d2 <- data.frame(y1=c(3,2,1),y2=c(6,5,4),y3=c(11,12,13))
mylist <- list(d1, d2)
par(mfrow=c(2,1))
# lapply on a subset of columns
lapply(mylist, function(x) plot(x$y2, x$y3))

You don't need a for loop to get their data points. You can call the column by their column names.
# a toy dataframe
d <- data.frame(A = 1:20, B = sample(c(FALSE, TRUE), 20, replace = TRUE),
C = LETTERS[1:20], D = rnorm(20, 0, 1))
col_names <- c("A", "B", "D") # names of columns I want to get
d[,col_names] # returns a dataset with the values of the columns you want

Here is a solution to your problem using a for loop:
# a toy dataframe
mylist <- list(dat1 = data.frame(A = 1:20, B = LETTERS[1:20]),
dat2 = data.frame(A = 21:40, B = LETTERS[1:20]),
dat3 = data.frame(A = 41:60, B = LETTERS[1:20]))
col_names <- c("A") # name of columns I want to get
for (i in 1:length(mylist)){
# you can do whatever you want with what is returned;
# here I am just print them out
print(names(mylist)[i]) # name of the data frame
print(mylist[[i]][,col_names]) # values in Column A
}

I think the simplest answer to your question is to use double brackets.
for (i in 1:length(my.list)) {
print(my.list[[i]]$column)
}
That works assuming all of the columns in your list of data frames have the same names. You could also call the position of the column in the data frame if you wanted.
Yes, lapply can be more elegant, but in some situations a for loop makes more sense.

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