I have the following dataset
structure(list(Var1 = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L,
2L), .Label = c("0", "1"), class = "factor"), Var2 = structure(c(1L,
1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("congruent", "incongruent"
), class = "factor"), Var3 = structure(c(1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L), .Label = c("spoken", "written"), class = "factor"),
Freq = c(8L, 2L, 10L, 2L, 10L, 2L, 10L, 2L)), class = "data.frame", row.names = c(NA,
-8L))
I would like to add another column reporting sum of coupled subsequent rows. Thus the final result would look like this:
I have proceeded like this
Table = as.data.frame(table(data_1$unimodal,data_1$cong_cond, data_1$presentation_mode)) %>%
mutate(Var1 = factor(Var1, levels = c('0', '1')))
row = Table %>% #is.factor(Table$Var1)
summarise(across(where(is.numeric),
~ .[Var1 == '0'] + .[Var1 == '1'],
.names = "{.col}_sum"))
column = c(rbind(row$Freq_sum,rep(NA, 4)))
Table$column = column
But I am looking for the quickest way possible with no scripting separated codes. Here I have used the dplyr package, but if you might know possibly suggest some other ways with map(), for loop, and or the method you deem as the best, please just let me know.
This should do:
df$column <-
rep(colSums(matrix(df$Freq, 2)), each=2) * c(1, NA)
If you are fine with no NAs in the dataframe, you can
df %>%
group_by(Var2, Var3) %>%
mutate(column = sum(Freq))
# A tibble: 8 × 5
# Groups: Var2, Var3 [4]
Var1 Var2 Var3 Freq column
<fct> <fct> <fct> <int> <int>
1 0 congruent spoken 8 10
2 1 congruent spoken 2 10
3 0 incongruent spoken 10 12
4 1 incongruent spoken 2 12
5 0 congruent written 10 12
6 1 congruent written 2 12
7 0 incongruent written 10 12
8 1 incongruent written 2 12
Related
I have data where each row represents a household, and I would like to have one row per individual in the different households.
The data looks similar to this:
df <- data.frame(village = rep("aaa",5),household_ID = c(1,2,3,4,5),name_1 = c("Aldo","Giovanni","Giacomo","Pippo","Pippa"),outcome_1 = c("yes","no","yes","no","no"),name_2 = c("John","Mary","Cindy","Eva","Doron"),outcome_2 = c("yes","no","no","no","no"))
I would still like to keep the wide format of the data, just with one individual (and related outcome variables) per row. I could find examples that tell how to do the opposite, going from individual to grouped data using dcast, but I could not find examples of this problem I am facing now.
I have tried with melt
reshape2::melt(df, id.vars = "household_ID")
but I get a long format data.
Any suggestions welcome...
Thank you
Use pivot_longer() in tidyr, and set ".value" in names_to to indicate new column names from the pattern of the original column names.
library(tidyr)
df %>%
pivot_longer(-c(village, household_ID),
names_to = c(".value", "n"),
names_sep = "_")
# # A tibble: 10 x 5
# village household_ID n name outcome
# <fct> <dbl> <chr> <fct> <fct>
# 1 aaa 1 1 Aldo yes
# 2 aaa 1 2 John yes
# 3 aaa 2 1 Giovanni no
# 4 aaa 2 2 Mary no
# 5 aaa 3 1 Giacomo yes
# 6 aaa 3 2 Cindy no
# 7 aaa 4 1 Pippo no
# 8 aaa 4 2 Eva no
# 9 aaa 5 1 Pippa no
# 10 aaa 5 2 Doron no
Data
df <- structure(list(village = structure(c(1L, 1L, 1L, 1L, 1L), .Label = "aaa", class = "factor"),
household_ID = c(1, 2, 3, 4, 5), name_1 = structure(c(1L,
3L, 2L, 5L, 4L), .Label = c("Aldo", "Giacomo", "Giovanni",
"Pippa", "Pippo"), class = "factor"), outcome_1 = structure(c(2L,
1L, 2L, 1L, 1L), .Label = c("no", "yes"), class = "factor"),
name_2 = structure(c(4L, 5L, 1L, 3L, 2L), .Label = c("Cindy",
"Doron", "Eva", "John", "Mary"), class = "factor"), outcome_2 = structure(c(2L,
1L, 1L, 1L, 1L), .Label = c("no", "yes"), class = "factor")), class = "data.frame", row.names = c(NA, -5L))
This question already has answers here:
How to reshape data from long to wide format
(14 answers)
Closed 3 years ago.
I have a data frame sex(male & female), age(child & adult), survive(yes & no) and frequency. How can I create a cross tab of sex and age?
sex age survive freq
male child yes 4
male adult yes 0
female child yes 6
female adult yes 3
male child no 1
male adult no 0
female child no 2
female adult no 1
I think you are looking for reshaping your data using pivot_wider from tidyr:
library(tidyr)
df %>% pivot_wider(., names_from = age, values_from = freq)
# A tibble: 4 x 4
sex survive child adult
<fct> <fct> <int> <int>
1 male yes 4 0
2 female yes 6 3
3 male no 1 0
4 female no 2 1
or
library(tidyr)
df %>% pivot_wider(., names_from = c(age, survive), values_from = freq)
# A tibble: 2 x 5
sex child_yes adult_yes child_no adult_no
<fct> <int> <int> <int> <int>
1 male 4 0 1 0
2 female 6 3 2 1
Is it what you are looking for ? If not, can you provide the expected outcome ?
Data
df = structure(list(sex = structure(c(2L, 2L, 1L, 1L, 2L, 2L, 1L,
1L), .Label = c("female", "male"), class = "factor"), age = structure(c(2L,
1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c("adult", "child"), class = "factor"),
survive = structure(c(2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L), .Label = c("no",
"yes"), class = "factor"), freq = c(4L, 0L, 6L, 3L, 1L, 0L,
2L, 1L)), class = "data.frame", row.names = c(NA, -8L))
I have trouble combining slice and map.
I am interested of doing something similar to this; which is, in my case, transforming a compact person-period file to a long (sequential) person-period one. However, because my file is too big, I need to split the data first.
My data look like this
group id var ep dur
1 A 1 a 1 20
2 A 1 b 2 10
3 A 1 a 3 5
4 A 2 b 1 5
5 A 2 b 2 10
6 A 2 b 3 15
7 B 1 a 1 20
8 B 1 a 2 10
9 B 1 a 3 10
10 B 2 c 1 20
11 B 2 c 2 5
12 B 2 c 3 10
What I need is simply this (answer from this)
library(dplyr)
dt %>% slice(rep(1:n(),.$dur))
However, I am interested in introducing a split(.$group).
How I am suppose to do so ?
dt %>% split(.$group) %>% map_df(slice(rep(1:n(),.$dur)))
Is not working for example.
My desired output is the same as dt %>% slice(rep(1:n(),.$dur))
which is
group id var ep dur
1 A 1 a 1 20
2 A 1 a 1 20
3 A 1 a 1 20
4 A 1 a 1 20
5 A 1 a 1 20
6 A 1 a 1 20
7 A 1 a 1 20
8 A 1 a 1 20
9 A 1 a 1 20
10 A 1 a 1 20
.....
But I need to split this operation because the file is too big.
data
dt = structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L), .Label = c("A", "B"), class = "factor"),
id = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L,
2L, 2L), .Label = c("1", "2"), class = "factor"), var = structure(c(1L,
2L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 3L, 3L, 3L), .Label = c("a",
"b", "c"), class = "factor"), ep = structure(c(1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("1", "2",
"3"), class = "factor"), dur = c(20, 10, 5, 5, 10, 15, 20,
10, 10, 20, 5, 10)), .Names = c("group", "id", "var", "ep",
"dur"), row.names = c(NA, -12L), class = "data.frame")
map takes two arguments: a vector/list in .x and a function in .f. It then applies .f on all elements in .x.
The function you are passing to map is not formatted correctly. Try this:
f <- function(x) x %>% slice(rep(1:n(), .$dur))
dt %>%
split(.$group) %>%
map_df(f)
You could also use it like this:
dt %>%
split(.$group) %>%
map_df(slice, rep(1:n(), dur))
This time you directly pass the slice function to map with additional parameters.
I'm not quite sure what your desired final output is, but you could use tidyr to nest the data that you want to repeat and a simple function to expand levels of your nested data, very similar to Tutuchan's answer.
expand_df <- function(df, repeats) {
df %>% slice(rep(1:n(), repeats))
}
dt %>%
tidyr::nest(var:ep) %>%
mutate(expanded = purrr::map2(data, dur, expand_df)) %>%
select(-data) %>%
tidyr::unnest()
Tutuchan's answer gives exactly the same output as your original approach - is that what you were looking for? I don't know if it will have any advantage over your original method.
Because I am working on a very large dataset, I need to slice my dataset by groups in order to pursue my computations.
I have a person-period (melt) dataset that looks like this
group id var time
1 A 1 a 1
2 A 1 b 2
3 A 1 a 3
4 A 2 b 1
5 A 2 b 2
6 A 2 b 3
7 B 1 a 1
8 B 1 a 2
9 B 1 a 3
10 B 2 c 1
11 B 2 c 2
12 B 2 c 3
I need to do this simple transformation
library(reshape2)
library(dplyr)
dt %>% dcast(group + id ~ time, value.var = 'var')
In order to get
group id 1 2 3
1 A 1 a b a
2 A 2 b b b
3 B 1 a a a
4 B 2 c c c
So far, so good.
However, because my database is too big, I need to do this separately for each different groups, such as
a = dt %>% filter(group == 'A') %>% dcast(group + id ~ time, value.var ='var')
b = dt %>% filter(group == 'B') %>% dcast(group + id ~ time, value.var = 'var')
bind_rows(a,b)
My problem is that I would like to avoid doing it by hand. I mean, having to store separately each groups, a = ..., b = ..., c = ..., and so on
Any idea how I could have a single pipe stream that would separate each group, compute the transformation and put it back together in a dataframe ?
dt = structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L, 2L, 2L), .Label = c("A", "B"), class = "factor"),
id = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L), .Label = c("1", "2"), class = "factor"), var = structure(c(1L,
2L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 3L, 3L, 3L), .Label = c("a",
"b", "c"), class = "factor"), time = structure(c(1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("1",
"2", "3"), class = "factor")), .Names = c("group", "id",
"var", "time"), row.names = c(NA, -12L), class = "data.frame")
Package purrr can be useful for working with lists. First split the dataset by group and then use map_df to dcast each list but return everything in a single data.frame.
library(purrr)
dt %>%
split(.$group) %>%
map_df(~dcast(.x, group + id ~ time, value.var = "var"))
group id 1 2 3
1 A 1 a b a
2 A 2 b b b
3 B 1 a a a
4 B 2 c c c
lapply is your friend here:
do.call(rbind, lapply(unique(dt$Group), function(grp, dt){
dt %>% filter(Group == grp) %>% dcast(group + id ~ time, value.var = "var")
}, dt = dt))
Using the following dataframe I would like to group the data by replicate and group and then calculate a ratio of treatment values to control values.
structure(list(group = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L), .Label = c("case", "controls"), class = "factor"), treatment = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "EPA", class = "factor"),
replicate = structure(c(2L, 4L, 3L, 1L, 2L, 4L, 3L, 1L), .Label = c("four",
"one", "three", "two"), class = "factor"), fatty_acid_family = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "saturated", class = "factor"),
fatty_acid = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "14:0", class = "factor"),
quant = c(6.16, 6.415, 4.02, 4.05, 4.62, 4.435, 3.755, 3.755
)), .Names = c("group", "treatment", "replicate", "fatty_acid_family",
"fatty_acid", "quant"), class = "data.frame", row.names = c(NA,
-8L))
I have tried using dplyr as follows:
group_by(dataIn, replicate, group) %>% transmute(ratio = quant[group=="case"]/quant[group=="controls"])
but this results in Error: incompatible size (%d), expecting %d (the group size) or 1
Initially I thought this might be because I was trying to create 4 ratios from a df 8 rows deep and so I thought summarise might be the answer (collapsing each group to one ratio) but that doesn't work either (my understanding is a shortcoming).
group_by(dataIn, replicate, group) %>% summarise(ratio = quant[group=="case"]/quant[group=="controls"])
replicate group ratio
1 four case NA
2 four controls NA
3 one case NA
4 one controls NA
5 three case NA
6 three controls NA
7 two case NA
8 two controls NA
I would appreciate some advice on where I'm going wrong or even if this can be done with dplyr.
Thanks.
You can try:
group_by(dataIn, replicate) %>%
summarise(ratio = quant[group=="case"]/quant[group=="controls"])
#Source: local data frame [4 x 2]
#
# replicate ratio
#1 four 1.078562
#2 one 1.333333
#3 three 1.070573
#4 two 1.446449
Because you grouped by replicate and group, you could not access data from different groups at the same time.
#talat's answer solved for me. I created a minimal reproducible example to help my own understanding:
df <- structure(list(a = c("a", "a", "b", "b", "c", "c", "d", "d"),
b = c(1, 2, 1, 2, 1, 2, 1, 2), c = c(22, 15, 5, 0.2, 107,
6, 0.2, 4)), row.names = c(NA, -8L), class = c("tbl_df",
"tbl", "data.frame"))
# a b c
# 1 a 1 22.0
# 2 a 2 15.0
# 3 b 1 5.0
# 4 b 2 0.2
# 5 c 1 107.0
# 6 c 2 6.0
# 7 d 1 0.2
# 8 d 2 4.0
library(dplyr)
df %>%
group_by(a) %>%
summarise(prop = c[b == 1] / c[b == 2])
# a prop
# 1 a 1.466667
# 2 b 25.000000
# 3 c 17.833333
# 4 d 0.050000