Im sure someone has a smart solution for this problem:
I have a dataframe like so:
A <- c("name1", "name2", "name3", "name4", "name5", "name6")
B <- c(10, 8, 7, 3, -1, -2)
C <- c(8, 3, -1, -10, -2, -2)
df <- data.frame(A, B, C)
df
A B C
1 name1 10 8
2 name2 8 3
3 name3 7 -1
4 name4 3 -10
5 name5 -1 -2
6 name6 -2 -2
I want to obtain four values, by counting the rows if certain conditions are met:
I want to count the number of rows in this dataframe where both B and C are negative integers (>0) -- for this example that would be "2"
I want to count the number of rows in this dataframe where both B and C are positive integers (<0)-- for this example that would be "2"
I want to count the number of rows in this dataframe where B is a negative integer (>0) and C is positive -- for this example that would be "0"
I want to count the number of rows in this dataframe where B is a postive integer and C is negative) -- for this example that would be "2"
Im suspecting that this can be achieved with some sort of If/Else statement, combined with the "table(sign..." command?
Try this:
library(dplyr)
df_count <- df %>% summarise(con1 = sum(B < 0 & C < 0),
con2 = sum(B > 0 & C > 0),
con3 = sum(B < 0 & C > 0),
con4 = sum(B > 0 & C < 0))
df_count
con1 con2 con3 con4
2 2 0 2
We can use count after creating a column with interaction on the sign
library(dplyr)
df %>%
transmute(con = factor(interaction(sign(B), sign(C), sep=" "),
levels = c('1 1', '1 -1', '-1 1', '-1 -1'))) %>%
count(con, .drop = FALSE)
# con n
#1 1 1 2
#2 1 -1 2
#3 -1 1 0
#4 -1 -1 2
Related
I have a data frame - in which I have a column with a lengthy string separated by _. Now I am interested in counting the patterns and several possible combinations from the long string. In the use case I provided below, you can find that I would like to count the occurrence of events A and B but not anything else.
If A and B repeat like A_B or B_A alone or if they repeats itself n number of times, I want to count them and also if there are several occurrences of those combinations.
Example data frame:
participant <- c("A", "B", "C")
trial <- c(1,1,2)
string_pattern <- c("A_B_A_C_A_B", "B_A_B_A_C_D_A_B", "A_B_C_A_B")
df <- data.frame(participant, trial, string_pattern)
Expected output:
participant trial string_pattern A_B B_A A_B_A B_A_B B_A_B_A
1. A 1 A_B_A_C_A_B 2 1 1 0 0
2. B 1 B_A_B_A_C_D_A_B 2 2 1 1 1
3. C 2 A_B_C_A_B 2 0 0 0 0
My code:
revised_df <- df%>%
dplyr::mutate(A_B = stringr::str_count(string_pattern, "A_B"),
B_A = stringr::str_count(string_pattern, "B_A"),
B_A_B = string::str_count(string_pattern, "B_A_B"))
My approach gets complicated as the number of combinations increases. Hence, looking for a better solution.
You could write a function to solve this:
m <- function(s){
a <- seq(nchar(s)-1)
start <- rep(a, rev(a))
stop <- ave(start, start, FUN = \(x)seq_along(x)+x)
b <- substring(s, start, stop)
gsub('(?<=\\B)|(?=\\B)', '_', b, perl = TRUE)
}
n <- function(x){
names(x) <- x
a <- strsplit(gsub("_", '', gsub("_[^AB]+_", ':', x)), ':')
b <- t(table(stack(lapply(a, \(y)unlist(sapply(y, m))))))
data.frame(pattern=x, as.data.frame.matrix(b), row.names = NULL)
}
n(string_pattern)
pattern A_B A_B_A B_A B_A_B B_A_B_A
1 A_B_A_C_A_B 2 1 1 0 0
2 B_A_B_A_C_D_A_B 2 1 2 1 1
3 A_B_C_A_B 2 0 0 0 0
Try: This checks each string row for current column name
library(dplyr)
df |>
mutate(A_B = 0, B_A = 0, A_B_A = 0, B_A_B = 0, B_A_B_A = 0) |>
mutate(across(A_B:B_A_B_A, ~ str_count(string_pattern, cur_column())))
participant trial string_pattern A_B B_A A_B_A B_A_B B_A_B_A
1 A 1 A_B_A_C_A_B 2 1 1 0 0
2 B 1 B_A_B_A_C_D_A_B 2 2 1 1 1
3 C 2 A_B_C_A_B 2 0 0 0 0
Could anyone explain how to change the negative values in the below dataframe?
we have been asked to create a data structure to get the below output.
# > df
# x y z
# 1 a -2 3
# 2 b 0 4
# 3 c 2 -5
# 4 d 4 6
Then we have to use control flow operators and/or vectorisation to multiply only the negative values by 10.
I tried so many different ways but cannot get this to work. I get an error when i try to use a loop and because of the letters.
Create indices of the negative values and multiply by 10, i.e.
i1 <- which(df < 0, arr.ind = TRUE)
df[i1] <- as.numeric(df[i1]) * 10
# x y z
#1 a -20 3
#2 b 0 4
#3 c 2 -50
#4 d 4 6
First find out the numeric columns of the dataframe and multiply the negative values by 10.
cols <- sapply(df, is.numeric)
#Multiply negative values by 10 and positive with 1
df[cols] <- df[cols] * ifelse(sign(df[cols]) == -1, 10, 1)
df
# x y z
#1 a -20 3
#2 b 0 4
#3 c 2 -50
#4 d 4 6
Using dplyr -
library(dplyr)
df <- df %>% mutate(across(where(is.numeric), ~. * ifelse(sign(.) == -1, 10, 1)))
I have a list of elemental compositions, each element in it's own row. Sometimes these elements have a zero.
C H N O S
1 5 5 0 0 0
2 6 4 1 0 1
3 4 6 2 1 0
I need to combine them so that they read, e.g. C5H5, C6H4NS, C4H6N2O.
This means that for any element of value "1" I should only take the column name, and for anything with value 0, the column should be skipped altogether.
I'm not really sure where to start here. I could add a new column to make it easier to read across the columns, e.g.
c C h H n N o O s S
1 C 5 H 5 N 0 O 0 S 0
2 C 6 H 4 N 1 O 0 S 1
3 C 4 H 6 N 2 O 1 S 0
This way, I just need the output to be a single string, but I need to ignore any zero values, and drop the one after the element name.
And here a base R solution:
df = read.table(text = "
C H N O S
5 5 0 0 0
6 4 1 0 1
4 6 2 1 0
", header=T)
apply(df, 1, function(x){return(gsub('1', '', paste0(colnames(df)[x > 0], x[x > 0], collapse='')))})
[1] "C5H5" "C6H4NS" "C4H6N2O"
paste0(colnames(df)[x > 0], x[x > 0], collapse='') pastes together the column names where the row values are bigger than zero. gsub then removes the ones. And apply does this for each row in the data frame.
Here's a tidyverse solution that uses some reshaping:
df = read.table(text = "
C H N O S
5 5 0 0 0
6 4 1 0 1
4 6 2 1 0
", header=T)
library(tidyverse)
df %>%
mutate(id = row_number()) %>% # add row id
gather(key, value, -id) %>% # reshape data
filter(value != 0) %>% # remove any zero rows
mutate(value = ifelse(value == 1, "", value)) %>% # replace 1 with ""
group_by(id) %>% # for each row
summarise(v = paste0(key, value, collapse = "")) # create the string value
# # A tibble: 3 x 2
# id v
# <int> <chr>
# 1 1 C5H5
# 2 2 C6H4NS
# 3 3 C4H6N2O
Assume that the input matrix m is as given reproducibly in the Note at the end -- convert it to a matrix if it is a data frame using as.matrix.
Now create a matrix the same shape as m with just the letters so now lets contains the letters and m contains the numbers. Then paste the letters and numbers together and replace those cells for which the number is zero with the empty string. Also replace any cells for which the number is 1 with just the letter. Finally paste each row together. No packages are used and no loops or *apply are used.
lets <- t(replace(t(m), TRUE, colnames(m)))
mm <- paste0(lets, m)
mm <- replace(mm, m == 0, "")
mm <- ifelse(m == 1, lets, mm)
do.call("paste0", as.data.frame(mm))
## [1] "C5H5" "C6H4NS" "C4H6N2O"
Note
the input matrix m in reproducible form is assumed to be:
m <- matrix(c(5, 6, 4, 5, 4, 6, 0, 1, 2, 0, 0, 1, 0, 1, 0), 3, 5,
dimnames = list(NULL, c("C", "H", "N", "O", "S")))
Another idea that avoids the apply with margin 1,
gsub('1', '', sapply(split(df, 1:nrow(df)), function(i)
paste(paste0(names(i)[i != 0], i[i != 0]), collapse = '')))
# 1 2 3
# "C5H5" "C6H4NS" "C4H6N2O"
Another option
library(dplyr)
#Get indices of all non-zero numbers in the dataframe
inds <- which(df!=0, arr.ind = TRUE)
#Create a dataframe with row index, column index and value at that position
vals <- data.frame(inds, val = df[inds])
#For each row paste the name of the column and value together and then replace 1
vals %>%
group_by(row) %>%
summarise(chemical = paste0(names(df)[col], val,collapse = "")) %>%
mutate(chemical = gsub("[1]", "", chemical))
# row chemical
# <int> <chr>
#1 1 C5H5
#2 2 C6H4NS
#3 3 C4H6N2O
Any suggestion to select the columns of the row when value =1 and the sum columns values =1. it means that I will just select unique values, non-shared with the other individuals.
indv. X Y Z W T J
A 1 0 1 0 0 1
B 0 1 1 0 0 0
C 0 0 1 1 0 0
D 0 0 1 0 1 0
A: X, J
B: Y
C: W
D: T
Here you go! A solution in base r.
First we simulate your data, a data.frame with named rows and columns.
You can use sapply() to loop over the column indices.
A for-loop over the column indices will achieve the same thing.
Finally, save the results in a data.frame however you want.
# Simulate your example data
df <- data.frame(matrix(c(1, 0, 1, 0, 0, 1,
0, 1, 1, 0, 0, 0,
0, 0, 1, 1, 0, 0,
0, 0, 1, 0, 1, 0), nrow = 4, byrow = T))
# Names rows and columns accordingly
names(df) <- c("X", "Y", "Z", "W", "T", "J")
rownames(df) <- c("A", "B","C", "D")
> df
X Y Z W T J
A 1 0 1 0 0 1
B 0 1 1 0 0 0
C 0 0 1 1 0 0
D 0 0 1 0 1 0
Then we select columns where the sum == 1- columns with unique values.
For every one of these columns, we find the row of this value.
# Select columns with unique values (if sum of column == 1)
unique.cols <- which(colSums(df) == 1)
# For every one of these columns, select the row where row-value==1
unique.rows <- sapply(unique.cols, function(x) which(df[, x] == 1))
> unique.cols
X Y W T J
1 2 4 5 6
> unique.rows
X Y W T J
1 2 3 4 1
The rows are not named correctly yet (they are still the element named of unique.cols). So we reference the rownames of df to get the rownames.
# Data.frame of unique values
# Rows and columns in separate columns
df.unique <- data.frame(Cols = unique.cols,
Rows = unique.rows,
Colnames = names(unique.cols),
Rownames = rownames(df)[unique.rows],
row.names = NULL)
The result:
df.unique
Cols Rows Colnames Rownames
1 1 1 X A
2 2 2 Y B
3 4 3 W C
4 5 4 T D
5 6 1 J A
Edit:
This is how you could summarise the values per row using dplyr.
library(dplyr)
df.unique %>% group_by(Rownames) %>%
summarise(paste(Colnames, collapse=", "))
# A tibble: 4 x 2
Rownames `paste(Colnames, collapse = ", ")`
<fct> <chr>
1 A X, J
2 B Y
3 C W
4 D T
One idea is to use rowwise apply to find the columns with 1, after we filter out the columns with sum != to 1, i.e.
apply(df[colSums(df) == 1], 1, function(i) names(df[colSums(df) == 1])[i == 1])
$A
[1] "X" "J"
$B
[1] "Y"
$C
[1] "W"
$D
[1] "T"
You can play around with the output to get it to desired state, i.e.
apply(df[colSums(df) == 1], 1, function(i) toString(names(df[colSums(df) == 1])[i == 1]))
# A B C D
#"X, J" "Y" "W" "T"
Or
data.frame(cols = apply(df[colSums(df) == 1], 1, function(i) toString(names(df[colSums(df) == 1])[i == 1])))
# cols
#A X, J
#B Y
#C W
#D T
Here is an option with tidyverse. We gather the dataset to 'long' format, grouped by 'key', fiter the rows where 'val' is 1 and the sum of 'val is 1, grouped by 'indv.', summarise the 'key' by pasteing the elements together
library(dplyr)
library(tidyr)
gather(df1, key, val, -indv.) %>%
group_by(key) %>%
filter(sum(val) == 1, val == 1) %>%
group_by(indv.) %>%
summarise(key = toString(key))
# A tibble: 4 x 2
# indv. key
# <chr> <chr>
#1 A X, J
#2 B Y
#3 C W
#4 D T
I have a dataframe d like this:
ID Value1 Value2 Value3
1 20 25 0
2 2 0 0
3 15 32 16
4 0 0 0
What I would like to do is calculate the variance for each person (ID), based only on non-zero values, and to return NA where this is not possible.
So for instance, in this example the variance for ID 1 would be var(20, 25),
for ID 2 it would return NA because you can't calculate a variance on just one entry, for ID 3 the var would be var(15, 32, 16) and for ID 4 it would again return NULL because it has no numbers at all to calculate variance on.
How would I go about this? I currently have the following (incomplete) code, but this might not be the best way to go about it:
len=nrow(d)
variances = numeric(len)
for (i in 1:len){
#get all nonzero values in ith row of data into a vector nonzerodat here
currentvar = var(nonzerodat)
Variances[i]=currentvar
}
Note this is a toy example, but the dataset I'm actually working with has over 40 different columns of values to calculate variance on, so something that easily scales would be great.
Data <- data.frame(ID = 1:4, Value1=c(20,2,15,0), Value2=c(25,0,32,0), Value3=c(0,0,16,0))
var_nonzero <- function(x) var(x[!x == 0])
apply(Data[, -1], 1, var_nonzero)
[1] 12.5 NA 91.0 NA
This seems overwrought, but it works, and it gives you back an object with the ids attached to the statistics:
library(reshape2)
library(dplyr)
variances <- df %>%
melt(., id.var = "id") %>%
group_by(id) %>%
summarise(variance = var(value[value!=0]))
Here's the toy data I used to test it:
df <- data.frame(id = seq(4), X1 = c(3, 0, 1, 7), X2 = c(10, 5, 0, 0), X3 = c(4, 6, 0, 0))
> df
id X1 X2 X3
1 1 3 10 4
2 2 0 5 6
3 3 1 0 0
4 4 7 0 0
And here's the result:
id variance
1 1 14.33333
2 2 0.50000
3 3 NA
4 4 NA