R. lapply multinomial test to list of dataframes
I have a data frame A, which I split into a list of 100 data frames, each having 3 rows (In my real data each data frame has 500 rows). Here I show A with 2 elements of the list (row1-row3; row4-row6):
A <- data.frame(n = c(0, 1, 2, 0, 1, 2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(24878, 33605, 12100 , 25899, 34777, 13765))
# This is the list:
nest <- split(A, rep(1:2, each = 3))
I want to apply the multinomial test to each of these data frames and extract the p-value of each test. So far I have done this:
library(EMT)
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = FALSE, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
lapply(nest, fun)
However, I get:
"Error in multinomial.test(x$counts_set, prob = x$norm_genome, useChisq = F, :
Observations have to be stored in a vector, e.g. 'observed <- c(5,2,1)'"
Does anyone have a smarter way of doing this?
r dataframe lapply multinomial
add a comment |
I have a data frame A, which I split into a list of 100 data frames, each having 3 rows (In my real data each data frame has 500 rows). Here I show A with 2 elements of the list (row1-row3; row4-row6):
A <- data.frame(n = c(0, 1, 2, 0, 1, 2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(24878, 33605, 12100 , 25899, 34777, 13765))
# This is the list:
nest <- split(A, rep(1:2, each = 3))
I want to apply the multinomial test to each of these data frames and extract the p-value of each test. So far I have done this:
library(EMT)
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = FALSE, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
lapply(nest, fun)
However, I get:
"Error in multinomial.test(x$counts_set, prob = x$norm_genome, useChisq = F, :
Observations have to be stored in a vector, e.g. 'observed <- c(5,2,1)'"
Does anyone have a smarter way of doing this?
r dataframe lapply multinomial
I don't get an error when I run your code.
– Florian
Nov 10 at 5:31
add a comment |
I have a data frame A, which I split into a list of 100 data frames, each having 3 rows (In my real data each data frame has 500 rows). Here I show A with 2 elements of the list (row1-row3; row4-row6):
A <- data.frame(n = c(0, 1, 2, 0, 1, 2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(24878, 33605, 12100 , 25899, 34777, 13765))
# This is the list:
nest <- split(A, rep(1:2, each = 3))
I want to apply the multinomial test to each of these data frames and extract the p-value of each test. So far I have done this:
library(EMT)
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = FALSE, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
lapply(nest, fun)
However, I get:
"Error in multinomial.test(x$counts_set, prob = x$norm_genome, useChisq = F, :
Observations have to be stored in a vector, e.g. 'observed <- c(5,2,1)'"
Does anyone have a smarter way of doing this?
r dataframe lapply multinomial
I have a data frame A, which I split into a list of 100 data frames, each having 3 rows (In my real data each data frame has 500 rows). Here I show A with 2 elements of the list (row1-row3; row4-row6):
A <- data.frame(n = c(0, 1, 2, 0, 1, 2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(24878, 33605, 12100 , 25899, 34777, 13765))
# This is the list:
nest <- split(A, rep(1:2, each = 3))
I want to apply the multinomial test to each of these data frames and extract the p-value of each test. So far I have done this:
library(EMT)
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = FALSE, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
lapply(nest, fun)
However, I get:
"Error in multinomial.test(x$counts_set, prob = x$norm_genome, useChisq = F, :
Observations have to be stored in a vector, e.g. 'observed <- c(5,2,1)'"
Does anyone have a smarter way of doing this?
r dataframe lapply multinomial
r dataframe lapply multinomial
edited Nov 10 at 10:48
Florian
1,067817
1,067817
asked Nov 10 at 0:21
Lucas
3853616
3853616
I don't get an error when I run your code.
– Florian
Nov 10 at 5:31
add a comment |
I don't get an error when I run your code.
– Florian
Nov 10 at 5:31
I don't get an error when I run your code.
– Florian
Nov 10 at 5:31
I don't get an error when I run your code.
– Florian
Nov 10 at 5:31
add a comment |
2 Answers
2
active
oldest
votes
The results of split are created with names 1, 2 and so on. That's why x$count in fun cannot access it. To make it simpler, you can combine your splitted elements using the list function and then use lapply:
n <- c(0,1,2,0,1,2)
prob <- c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1)
count <- c(24878, 33605, 12100 , 25899, 34777, 13765)
A <- cbind.data.frame(n, prob, count)
nest = split(A,rep(1:2,each=3))
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = F, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
# Create a list of splitted elements
new_list <- list(nest$`1`, nest$`2`)
lapply(new_list, fun)
add a comment |
A solution with dplyr.
A = data.frame(n = c(0,1,2,0,1,2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(43, 42, 9, 74, 82, 9))
library(dplyr)
nest <- A %>%
mutate(pattern = rep(1:2,each=3)) %>%
group_by(pattern) %>%
dplyr::summarize(mn_pvals = multinomial.test(count, prob)$p.value)
nest
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
add a comment |
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2 Answers
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oldest
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2 Answers
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active
oldest
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oldest
votes
The results of split are created with names 1, 2 and so on. That's why x$count in fun cannot access it. To make it simpler, you can combine your splitted elements using the list function and then use lapply:
n <- c(0,1,2,0,1,2)
prob <- c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1)
count <- c(24878, 33605, 12100 , 25899, 34777, 13765)
A <- cbind.data.frame(n, prob, count)
nest = split(A,rep(1:2,each=3))
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = F, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
# Create a list of splitted elements
new_list <- list(nest$`1`, nest$`2`)
lapply(new_list, fun)
add a comment |
The results of split are created with names 1, 2 and so on. That's why x$count in fun cannot access it. To make it simpler, you can combine your splitted elements using the list function and then use lapply:
n <- c(0,1,2,0,1,2)
prob <- c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1)
count <- c(24878, 33605, 12100 , 25899, 34777, 13765)
A <- cbind.data.frame(n, prob, count)
nest = split(A,rep(1:2,each=3))
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = F, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
# Create a list of splitted elements
new_list <- list(nest$`1`, nest$`2`)
lapply(new_list, fun)
add a comment |
The results of split are created with names 1, 2 and so on. That's why x$count in fun cannot access it. To make it simpler, you can combine your splitted elements using the list function and then use lapply:
n <- c(0,1,2,0,1,2)
prob <- c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1)
count <- c(24878, 33605, 12100 , 25899, 34777, 13765)
A <- cbind.data.frame(n, prob, count)
nest = split(A,rep(1:2,each=3))
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = F, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
# Create a list of splitted elements
new_list <- list(nest$`1`, nest$`2`)
lapply(new_list, fun)
The results of split are created with names 1, 2 and so on. That's why x$count in fun cannot access it. To make it simpler, you can combine your splitted elements using the list function and then use lapply:
n <- c(0,1,2,0,1,2)
prob <- c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1)
count <- c(24878, 33605, 12100 , 25899, 34777, 13765)
A <- cbind.data.frame(n, prob, count)
nest = split(A,rep(1:2,each=3))
fun <- function(x)
multinomial.test(x$count,
prob=x$prob,
useChisq = F, MonteCarlo = TRUE,
ntrial = 100, # n of withdrawals accomplished
atOnce=100)
# Create a list of splitted elements
new_list <- list(nest$`1`, nest$`2`)
lapply(new_list, fun)
answered Nov 10 at 5:10
Vishesh Shrivastav
1,0462622
1,0462622
add a comment |
add a comment |
A solution with dplyr.
A = data.frame(n = c(0,1,2,0,1,2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(43, 42, 9, 74, 82, 9))
library(dplyr)
nest <- A %>%
mutate(pattern = rep(1:2,each=3)) %>%
group_by(pattern) %>%
dplyr::summarize(mn_pvals = multinomial.test(count, prob)$p.value)
nest
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
add a comment |
A solution with dplyr.
A = data.frame(n = c(0,1,2,0,1,2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(43, 42, 9, 74, 82, 9))
library(dplyr)
nest <- A %>%
mutate(pattern = rep(1:2,each=3)) %>%
group_by(pattern) %>%
dplyr::summarize(mn_pvals = multinomial.test(count, prob)$p.value)
nest
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
add a comment |
A solution with dplyr.
A = data.frame(n = c(0,1,2,0,1,2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(43, 42, 9, 74, 82, 9))
library(dplyr)
nest <- A %>%
mutate(pattern = rep(1:2,each=3)) %>%
group_by(pattern) %>%
dplyr::summarize(mn_pvals = multinomial.test(count, prob)$p.value)
nest
A solution with dplyr.
A = data.frame(n = c(0,1,2,0,1,2),
prob = c(0.4, 0.5, 0.1, 0.4, 0.5, 0.1),
count = c(43, 42, 9, 74, 82, 9))
library(dplyr)
nest <- A %>%
mutate(pattern = rep(1:2,each=3)) %>%
group_by(pattern) %>%
dplyr::summarize(mn_pvals = multinomial.test(count, prob)$p.value)
nest
answered Nov 10 at 6:58
paoloeusebi
641413
641413
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
add a comment |
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
Hi @paoloeusebi, your solution also works great, but Vishesh's posted a solution first. Thanks very much
– Lucas
Nov 10 at 22:19
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
The most important things are that you have two solutions and I have learned something new.
– paoloeusebi
Nov 10 at 22:21
Glad to hear that !
– Lucas
Nov 12 at 18:12
Glad to hear that !
– Lucas
Nov 12 at 18:12
add a comment |
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I don't get an error when I run your code.
– Florian
Nov 10 at 5:31