Rowwise maximum of multiple columns (ignoring null values) - sqlite

How to apply the max()-functions to multiple columns, ignoring NULL values
(in this context NULL is named NA).
My data:
# data
df <- data.frame( a = sample(5), b = sample(5) )
df[2:3,1] <- NA
dbWriteTable(db1, "df", df, overwrite = TRUE )
What I have tried
What I want:
(notice that the column max1 does not contain NA)
I was hoping there was a simple way to do this in SQLite, but may there is not?

df$max<-apply(X=df, MARGIN=1, FUN=max,na.rm=T)

Related

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

Is there an easy way to tell if many data frames stored in one list contain the same columns?

I have a list containing many data frames:
df1 <- data.frame(A = 1:5, B = 2:6, C = LETTERS[1:5])
df2 <- data.frame(A = 1:5, B = 2:6, C = LETTERS[1:5])
df3 <- data.frame(A = 1:5, C = LETTERS[1:5])
my_list <- list(df1, df2, df3)
I want to know if every data frame in this list contains the same columns (i.e., the same number of columns, all having the same names and in the same order).
I know that you can easily find column names of data frames in a list using lapply:
lapply(my_list, colnames)
Is there a way to determine if any differences in column names occur? I realize this is a complicated question involving pairwise comparisons.
You can avoid pairwise comparison by simply checking if the count of each column name is == length(my_list). This will simultaneously check for dim and names of you dataframe -
lapply(my_list, names) %>%
unlist() %>%
table() %>%
all(. == length(my_list))
[1] FALSE
In base R i.e. without %>% -
all(table(unlist(lapply(my_list, names))) == length(my_list))
[1] FALSE
or sightly more optimized -
!any(table(unlist(lapply(my_list, names))) != length(my_list))
Here's another base solution with Reduce:
!is.logical(
Reduce(function(x,y) if(identical(x,y)) x else FALSE
, lapply(my_list, names)
)
)
You could also account for same columns in a different order with
!is.logical(
Reduce(function(x,y) if(identical(x,y)) x else FALSE
, lapply(my_list, function(z) sort(names(z)))
)
)
As for what's going on, Reduce() accumulates as it goes through the list. At first, identical(names_df1, names_df2) are evaluated. If it's true, we want to have it return the same vector evaluated! Then we can keep using it to compare to other members of the list.
Finally, if everything evaluates as true, we get a character vector returned. Since you probably want a logical output, !is.logical(...) is used to turn that character vector into a boolean.
See also here as I was very inspired by another post:
check whether all elements of a list are in equal in R
And a similar one that I saw after my edit:
Test for equality between all members of list
We can use dplyr::bind_rows:
!any(is.na(dplyr::bind_rows(my_list)))
# [1] FALSE
Here is my answer:
k <- 1
output <- NULL
for(i in 1:(length(my_list) - 1)) {
for(j in (i + 1):length(my_list)) {
output[k] <- identical(colnames(my_list[[i]]), colnames(my_list[[j]]))
k <- k + 1
}
}
all(output)

Delete rows after a negative value in multiple data frames

I have multiple data frames which are individual sequences, consisting out the same columns. I need to delete all the rows after a negative value is encountered in the column "OnsetTime". So not the row of the negative value itself, but the row after that. All sequences have 16 rows in total.
I think it must be able by a loop, but I have no experience with loops in r and I have 499 data frames of which I am currently deleting the rows of a sequence one by one, like this:
sequence_6 <- sequence_6[-c(11:16), ]
sequence_7 <- sequence_7[-c(11:16), ]
sequence_9 <- sequence_9[-c(6:16), ]
Is there a faster way of doing this? An example of a sequence can be seen here example sequence
Ragarding this example, I want to delete row 7 to row 16
Data
Since the odd web configuration at work prevents me from accessing your data, I created three dataframes based on random numbers
set.seed(123); data_1 <- data.frame( value = runif(25, min = -0.1) )
set.seed(234); data_2 <- data.frame( value = runif(20, min = -0.1) )
set.seed(345); data_3 <- data.frame( value = runif(30, min = -0.1) )
First, you could create a list containing all your dataframes:
list_df <- list(data_1, data_2, data_3)
Now you can go through this list with a for loop. Since there are several steps, I find it convenient to use the package dplyr because it allows for a more readable notation:
library(dplyr)
for( i in 1:length(list_df) ){
min_row <-
list_df[[i]] %>%
mutate( id = row_number() ) %>% # add a column with row number
filter(value < 0) %>% # get the rows with negative values
summarise( min(id) ) %>% # get the first row number
as.numeric() # transform this value to a scalar (not a dataframe)
list_df[[i]] <- list_df[[i]] %>% slice(1:min_row) # get rows 1 to min_row
}
Hope it helps!
We can get the datasets into a list assuming that the object names start with 'sequence' followed by a - and one or more digits. Then use lapply to loop over the list and subset the rows based on the condition
lst1 <- lapply(mget(ls(pattern="^sequence_\\d+$")), function(x) {
i1 <- Reduce(`|`, lapply(x, `<`, 0))
#or use rowSums
#i1 <- rowSums(x < 0) > 0
i2 <- which(i1)[1]
x[seq(i2),]
}
)
data
set.seed(42)
sequence_6 <- as.data.frame(matrix(sample(-1:10, 16 *5, replace = TRUE), nrow = 16))
sequence_7 <- as.data.frame(matrix(sample(-2:10, 16 *5, replace = TRUE), nrow = 16))
sequence_9 <- as.data.frame(matrix(sample(-2:10, 16 *5, replace = TRUE), nrow = 16))

Replace value from dataframe column with value from keyvalue lookup

I want to replace certain values in a data frame column with values from a lookup table. I have the values in a list, stuff.kv, and many values are stored in the list (but some may not be).
stuff.kv <- list()
stuff.kv[["one"]] <- "thing"
stuff.kv[["two"]] <- "another"
#etc
I have a dataframe, df, which has multiple columns (say 20), with assorted names. I want to replace the contents of the column named 'stuff' with values from 'lookup'.
I have tried building various apply methods, but nothing has worked.
I built a function, which process a list of items and returns the mutated list,
stuff.lookup <- function(x) {
for( n in 1:length(x) ) {
if( !is.null( stuff.kv[[x[n]]] ) ) x[n] <- stuff.kv[[x[n]]]
}
return( x )
}
unlist(lapply(df$stuff, stuff.lookup))
The apply syntax is bedeviling me.
Since you made such a nice lookup table, You can just use it to change the values. No loops or apply needed.
## Sample Data
set.seed(1234)
DF = data.frame(stuff = sample(c("one", "two"), 8, replace=TRUE))
## Make the change
DF$stuff = unlist(stuff.kv[DF$stuff])
DF
stuff
1 thing
2 another
3 another
4 another
5 another
6 another
7 thing
8 thing
Below is a more general solution building on #G5W's answer as it doesn't cover the case where your original data frame has values that don't exist in the lookup table (which would result in length mismatch error):
library(dplyr)
stuff.kv <- list(one = "another", two = "thing")
df <- data_frame(
stuff = rep(c("one", "two", "three"), each = 3)
)
df <- df %>%
mutate(stuff = paste(stuff.kv[stuff]))

reference x's column in R's apply function

I have a df like this:
a <- c(4,5,3,5,1)
b <- c(8,9,7,3,5)
c <- c(6,7,5,4,3)
df <- data.frame(rbind(a,b,c))
I want a new df, df2, containing the difference between the values in each cell in rows a and b and the value in row c in their respective columns.
df2 would look like this:
a <- c(-2,-2,-2,1,-2)
b <- c(2,2,2,-1,2)
df2 <- data.frame(rbind(a,b))
Here is where I'm getting stuck:
df2 <- data.frame(apply(df,c(1,2),function(x) x - df[nrow(df),the col index of x]))
How do I reference the column index of x? Is there something like JavaScript's this?
We can do this easily by replicating the 3rd row to make the lengths equal before subtracting with the first two rows
out <- df[c("a", "b"),] - df["c",][col(df[c("a", "b"),])]
identical(df2, out)
#[1] TRUE
Or explicitly using rep
df[c("a", "b"),] - rep(unlist(df["c",]), each = 2)

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