enter image description hereI have the following vectors:
bid = c(1,5,10,20,30,40,50)
n = c(31,29,27,25,23,21,19)
yes = c(0,3,6,7,9,13,17)
no = n - yes
I have two questions, and I don't find any solutions for them, I would appreciate if someone can help me.
Q1: I want to write R code to create a two-column dataframe df. Column 1 has Bid,
where each Bid is repeated n times; Column 2 has c(rep(1,yes),rep(0,no) at
each bid.
Q2: Then when I have the data frame df, I want to write R codes to generate
(from df) vectors bid, n, yes, and no, again.
It is a bit unclear what you actually want. It is easier if you provide the desired result. Would this fit your Q1:
library(tidyverse)
bid = c(1,5,10,20,30,40,50)
n = c(31,29,27,25,23,21,19)
yes = c(0,3,6,7,9,13,17)
no = n - yes
df <- tibble(bid, yes, n, no = n -yes) %>% dplyr::select(- n) %>% pivot_longer(cols = c(yes, no)) %>% uncount(value) %>% mutate(yesno = ifelse(name == "yes", 1,0)) %>% dplyr::select(-name)
df2 <- df %>% group_by(bid) %>% table() %>% as.data.frame() %>% pivot_wider(id_cols = bid, names_from = yesno, values_from = Freq) %>% mutate(n = yes + no) %>% rename(no = `0`, yes = `1`)
bid <- df2$bid
n <- df2$n
yes <- df2$yes
I don't know what you mean for Q2, but for Q1 you could do this:
library(tidyverse)
pmap_dfr(list(bid, n, yes, no),
\(V1, V2, V3, V4) tibble(col1 = rep(V1, V2),
col2 = c(rep(1,V3),rep(0,V4))))
#> # A tibble: 175 x 2
#> col1 col2
#> <dbl> <dbl>
#> 1 1 0
#> 2 1 0
#> 3 1 0
#> 4 1 0
#> 5 1 0
#> 6 1 0
#> 7 1 0
#> 8 1 0
#> 9 1 0
#> 10 1 0
#> # ... with 165 more rows
EDIT:
For Q2, you can follow this:
library(tidyverse)
df <- pmap_dfr(list(bid, n, yes, no),
\(V1, V2, V3, V4) tibble(col1 = rep(V1, V2),
col2 = c(rep(1,V3),rep(0,V4))))
df2 <- df |>
count(col1, col2) |>
group_by(col1) |>
summarise(yes = sum(n[col2==1]),
n = sum(n))
bid2 <- df2$col1
n2 <- df2$n
yes2 <- df2$yes
no2 <- n2 - yes2
all.equal(c(bid, n, yes, no), c(bid2, n2, yes2, no2))
#> [1] TRUE
Related
I need to summarize one variable/column of a long table after aggregating (group_by()) by another variable/column, I need to have the summarized value by all values of other variables/columns.
Here is test data:
library(tidyverse)
set.seed(123)
Site <- str_c("S", 1:5)
Species <- str_c("Sps", 1:6)
print(Species_tbl <- bind_cols(Species = Species,
Exotic = rbinom(length(Species), 1, .3),
Migrant = rbinom(length(Species), 2, .3)))
Data_tbl <- expand.grid(Site = Site,
Species = Species) %>%
left_join(Species_tbl)
Data_tbl$Presence <- rbinom(nrow(Data_tbl), 1, .5)
And here is my best effort:
print(Data_tbl %>%
group_by(Site) %>%
summarise(N_sp = sum(Presence),
N_sp_Exo = sum(Presence[Exotic == 1]),
N_sp_Nat = sum(Presence[Exotic == 0]),
N_sp_M0 = sum(Presence[Migrant == 0]),
N_sp_M1 = sum(Presence[Migrant == 1]),
N_sp_M2 = sum(Presence[Migrant == 2])))
You can get the data in long format for your columns of interest c(Exotic, Migrant) and take sum of Presence columns for each unique column names and it's values. This can be merged with sum of each Site.
library(dplyr)
library(tidyr)
data1 <- Data_tbl %>%
group_by(Site) %>%
summarise(N_sp = sum(Presence))
data2 <- Data_tbl %>%
pivot_longer(cols = c(Exotic, Migrant)) %>%
group_by(Site, name, value) %>%
summarise(result = sum(Presence), .groups = "drop") %>%
pivot_wider(names_from = c(name, value), values_from = result)
inner_join(data1, data2, by = 'Site')
# Site N_sp Exotic_0 Exotic_1 Migrant_0 Migrant_1 Migrant_2
# <fct> <int> <int> <int> <int> <int> <int>
#1 S1 4 2 2 1 2 1
#2 S2 3 2 1 0 2 1
#3 S3 2 1 1 0 2 0
#4 S4 4 2 2 1 3 0
#5 S5 4 1 3 1 2 1
The answer has been divided in two steps for ease of readability. If you would like to do this in a single chain without creating temporary variables that can be done as well.
Let's say I had the following data frame, that was also altered to include counts of a,b, and c, based on whether or not they are classified by Z = 0 or 1
X <- (1:10)
Y<- c('a','b','a','c','b','b','a','a','c','c')
Z <- c(0,1,1,1,0,1,0,1,1,1)
test_df <- data.frame(X,Y,Z)
(the code below was provided by a stack exchange member, thank you!)
res <- test_df %>% group_by(Y,Z) %>% summarise(N=n()) %>%
pivot_wider(names_from = Z,values_from=N,
values_fill = 0)
How might I add a column on the right which would indicate the proportion of each of the letters for which z=1, out of all appearances of that letter? It would seem that a basic summary statement should work but I figure out how...
My expected output would be something like
Z=0 Z=1 PropZ=1
a 2 2 .5
b 1 2 .66
c 0 3 1
Perhaps this helps
library(dplyr)
library(tidyr)
test_df %>%
group_by(Y, Z) %>%
summarise(N = n(), .groups = 'drop') %>%
left_join(test_df %>%
group_by(Y) %>%
summarise(Prop = mean(Z == 1), .groups = 'drop')) %>%
pivot_wider(names_from = Z, values_from = N, values_fill = 0)
-output
# A tibble: 3 x 4
# Y Prop `0` `1`
# <chr> <dbl> <int> <int>
#1 a 0.5 2 2
#2 b 0.667 1 2
#3 c 1 0 3
test_df %>% group_by(Y) %>%
summarise( z0 = sum(Z == 0), z1 = sum(Z == 1) , PropZ = z1/n())
I am not sure if what is your expected output, but below might be some options
u <- xtabs(q ~ Y + Z, cbind(test_df, q = 1))
> u
Z
Y 0 1
a 2 2
b 1 2
c 0 3
or
> prop.table(u)
Z
Y 0 1
a 0.2 0.2
b 0.1 0.2
c 0.0 0.3
To calculate proportions of 1 for each letter you can use rowSums.
transform(res, prop_1 = `1`/rowSums(res[-1]))
In dplyr :
library(dplyr)
res %>%
ungroup %>%
mutate(prop_1 = `1`/rowSums(.[-1]))
# Y `0` `1` prop_1
# <chr> <int> <int> <dbl>
#1 a 2 2 0.5
#2 b 1 2 0.667
#3 c 0 3 1
I have a date frame with the fields PARTIDA (date), Operação (4 levels factor) and TT (numeric) .
I need to group by the PARTIDA column, pivot the Operation column counting to the frequency of each level and sum the TT column.
Like this:
I already tried something with dplyr but I could not get this result, can anyone help me?
Here's a two-step process that may get you what you want:
library(dplyr)
df <-
tibble(
partida = c("date1", "date2", "date3", "date1", "date2"),
operacao = c("D", "J", "C", "D", "M"),
tt = c(1, 2, 3, 4, 5)
)
tt_sums <-
df %>%
group_by(partida) %>%
count(wt = tt)
operacao_counts <-
df %>%
group_by(partida, operacao) %>%
count() %>%
ungroup() %>%
spread(operacao, n) %>%
mutate_if(is.numeric, replace_na, 0)
final_df <-
operacao_counts %>%
left_join(tt_sums, by = "partida")
> final_df
# A tibble: 3 x 6
partida C D J M n
<chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 date1 0 2 0 0 5
2 date2 0 0 1 1 7
3 date3 1 0 0 0 3
Similar to #cardinal40's answer but in one go as I try to limit the number of objects added to my environment when possible. Either answer will do the trick.
df %>%
group_by(partida) %>%
mutate(tt = sum(tt)) %>%
group_by(partida, operacao, tt) %>%
count() %>%
ungroup() %>%
spread(operacao, n) %>%
mutate_if(is.numeric, replace_na, 0)
My data = data.lab
data.lab <- data.frame(Name=c("A","e","b","c","d"),
bp =c( 12,12,11,12,11),
sugar = c(19,21,23,19,23))
I want to have only duplicate names with the reference
desired output
lab.data <- data.frame(Name=c("A","b","c","d"),
bp =c( 12,11,12,11),
sugar = c(19,23,19,23),
pair=c(1,1,2,2))
dub.data <- duplicated(data.lab) | duplicated(data.lab, fromLast = TRUE)
out.1=data.lab[dub.data, ]
this gives the duplicate data but i need a column as what are the duplicate pairs
With dplyr, you can do:
data.lab %>%
group_by(bp, sugar) %>%
filter(n() == 2) %>%
mutate(pair = seq_along(Name))
Name bp sugar pair
<fct> <dbl> <dbl> <int>
1 A 12 19 1
2 b 11 23 1
3 c 12 19 2
4 d 11 23 2
Or:
data.lab %>%
group_by(bp, sugar) %>%
filter(n() == 2) %>%
mutate(pair = row_number())
Or if there could be more than two pairs of duplicates:
data.lab %>%
group_by(bp, sugar) %>%
filter(n() > 1) %>%
mutate(pair = seq_along(Name))
Or:
data.lab %>%
group_by(bp, sugar) %>%
filter(n() > 1) %>%
mutate(pair = row_number())
Or to group by all variables except of "Name":
data.lab %>%
group_by_at(vars(-matches("(Name)"))) %>%
filter(n() > 1) %>%
mutate(pair = seq_along(Name))
Or:
data.lab %>%
group_by_at(vars(-matches("(Name)"))) %>%
filter(n() > 1) %>%
mutate(pair = row_number())
Continuing from your approach , we can use ave in base R
dat1 <- data.lab[duplicated(data.lab[c("bp", "sugar")]) |
duplicated(data.lab[c("bp", "sugar")], fromLast = TRUE) , ]
dat1$pair <- with(dat1, ave(Name, bp, sugar, FUN = seq_along))
dat1
# Name bp sugar pair
#1 A 12 19 1
#2 b 11 23 1
#3 c 12 19 2
#4 d 11 23 2
I have a question:
I have a dataset like this simple example:
df<-data.frame(ID=c("A","B","C","D"),
Score=c("15","16/18/19+2/6","3/+2","19/18/14"))
I want to end up with a dataset that has split the score numbers. I have a problem with the /+2 part. when it says "3/+2"it actually means: "3/3+2" which would finally give "3/5". So what I would like some help with, is to end up with a dataset like this:
ID Score
A 15
B 16/18/19/21/6
C 3/5
D 19/18/14
I already found out that I can then seperate the score by
df<-df %>%
mutate(Score = strsplit(as.character(ID), "/")) %>%
unnest(Score)
But I don't know how I can let the numbers duplicate and then sum when /+ occurs, could someone help me?
It could be probably solved in a more elegant way, but here is one possibility:
df %>%
mutate(Score = strsplit(as.character(Score), "/")) %>%
unnest() %>%
rowwise() %>%
mutate(Score = eval(parse(text = paste0(Score)))) %>%
group_by(ID) %>%
mutate(Score = paste0(Score, collapse = "/")) %>%
distinct()
ID Score
<fct> <chr>
1 A 15
2 B 16/18/21/6
3 C 3/5
4 D 19/18/14
Sample data:
df <- data.frame(ID=c("A","B","C","D"),
Score=c("15","16/18/19+2/6","3/3+2","19/18/14"))
It splits "Score" based on /, converts characters to expression by parse() and then transforms it back.
Using the data you provided and the pattern from #A. Suliman:
df %>%
mutate(Score = strsplit(gsub("(\\d+)/*\\+(\\d+)","\\1/\\1+\\2", Score), "/")) %>%
unnest() %>%
rowwise() %>%
mutate(Score = eval(parse(text = paste0(Score)))) %>%
group_by(ID) %>%
mutate(Score = paste0(Score, collapse = "/")) %>%
distinct()
ID Score
<fct> <chr>
1 A 15
2 B 16/18/19/21/6
3 C 3/5
4 D 19/18/14
We can use gsubfn to do this in a compact way
library(gsubfn)
library(tidyverse)
df %>%
mutate(Score = gsubfn("\\d+\\+\\d+", ~ eval(parse(text = x)), Score))
# ID Score
#1 A 15
#2 B 16/18/21/6
#3 C 3/5
#4 D 19/18/14
data
df <- data.frame(ID=c("A","B","C","D"),
Score=c("15","16/18/19+2/6","3/3+2","19/18/14"), stringsAsFactors = FALSE)
library(dplyr)
library(tidyr) #separate_rows, no need for unnest
df %>% rowwise()%>%
mutate(Score_upd=paste0(sapply(unlist(strsplit(gsub('(\\d+)/*\\+(\\d+)','\\1/\\1+\\2',Score),'/')),
function(x)eval(parse(text = x))),collapse = '/')) %>%
separate_rows(Score_upd,sep = '/')
#short version
df %>% mutate(Score=gsub('(\\d+)/*\\+(\\d+)','\\1/\\1+\\2',Score)) %>%
separate_rows(Score,sep='/') %>% rowwise() %>% mutate(Score=eval(parse(text=Score))) %>%
group_by(ID) %>% summarise(Score=paste0(Score,collapse = '/'))
# A tibble: 4 x 2
ID Score
<fct> <chr>
1 A 15
2 B 16/18/19/21/6
3 C 3/5
4 D 19/18/14
The main idea is using gsub to separate 2+3 correctly, e.g:
gsub('(\\d+)/*\\+(\\d+)','\\1/\\1+\\2','20/8/2+3') #/* means 0 or 1 occurence of / e.g, 19+2 and 3/+2.
[1] "20/8/2/2+3"
Then
valid_str <- gsub('(\\d+)/*\\+(\\d+)','\\1/\\1+\\2','20/8/2+3')
sapply(unlist(strsplit(valid_str,'/')),function(x) eval(parse(text=x)))
20 8 2 2+3
20 8 2 5
#OR
sapply(unlist(strsplit(valid_str,'/')),function(x) sum(as.numeric(unlist(strsplit(x,'\\+')))))
20 8 2 2+3
20 8 2 5