I am currently using cast on a melted table to calculate the total of each value at the combination of ID variables ID1 (row names) and ID2 (column headers), along with grand totals for each row using margins="grand_col".
c <- cast(m, ID1 ~ ID2, sum, margins="grand_col")
ID1 ID2a ID2b ID2c ID2d ID2e (all)
1 ID1a 6459695 885473 648019 453613 1777308 10224108
2 ID1b 7263529 1411355 587785 612730 2458672 12334071
3 ID1c 7740364 1253524 682977 886897 3559283 14123045
So far, so R-like.
Then I divide each cell by its row total to get a percentage of the total.
c[,2:6]<-c[,2:6] / c[,7]
This looks kludgy. Is there something I should be doing in cast or maybe in plyr to handle the percent of margin calculation in the first command?
Thanks,
Matt
Assuming your source table looks something like this:
dfm <- structure(list(ID1 = structure(c(1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L), .Label = c("ID1a", "ID1b", "ID1c"
), class = "factor"), ID2 = structure(c(1L, 1L, 1L, 2L,
2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L, 5L, 5L, 5L), .Label = c("ID2a",
"ID2b", "ID2c", "ID2d", "ID2e"), class = "factor"), value = c(6459695L,
7263529L, 7740364L, 885473L, 1411355L, 1253524L, 648019L, 587785L,
682977L, 453613L, 612730L, 886897L, 1777308L, 2458672L, 3559283L
)), .Names = c("ID1", "ID2", "value"), row.names = c(NA,
-15L), class = "data.frame")
> head(dfm)
ID1 ID2 value
1 ID1a ID2a 6459695
2 ID1b ID2a 7263529
3 ID1c ID2a 7740364
4 ID1a ID2b 885473
5 ID1b ID2b 1411355
6 ID1c ID2b 1253524
Using ddply first to calculate the percentages, and cast to present the data in the required format
library(reshape)
library(plyr)
df1 <- ddply(dfm, .(ID1), summarise, ID2 = ID2, pct = value / sum(value))
dfc <- cast(df1, ID1 ~ ID2)
dfc
ID1 ID2a ID2b ID2c ID2d ID2e
1 ID1a 0.6318101 0.08660638 0.06338147 0.04436700 0.1738350
2 ID1b 0.5888996 0.11442735 0.04765539 0.04967784 0.1993399
3 ID1c 0.5480662 0.08875735 0.04835905 0.06279786 0.2520195
Compared to your example, this is missing the row totals, these need to be added separately.
Not sure though, whether this solution is more elegant than the one you currently have.
Here is a one-liner using tapply and prop.table. It does not rely on any auxilliary packages:
prop.table(tapply(dfm$value, dfm[1:2], sum), 1)
giving:
ID2
ID1 ID2a ID2b ID2c ID2d ID2e
ID1a 0.6318101 0.08660638 0.06338147 0.04436700 0.1738350
ID1b 0.5888996 0.11442735 0.04765539 0.04967784 0.1993399
ID1c 0.5480662 0.08875735 0.04835905 0.06279786 0.2520195
or this which is even shorter:
prop.table( xtabs(value ~., dfm), 1 )
Related
I have the following dataset
structure(list(Var1 = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L,
2L), .Label = c("0", "1"), class = "factor"), Var2 = structure(c(1L,
1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("congruent", "incongruent"
), class = "factor"), Var3 = structure(c(1L, 1L, 1L, 1L, 2L,
2L, 2L, 2L), .Label = c("spoken", "written"), class = "factor"),
Freq = c(8L, 2L, 10L, 2L, 10L, 2L, 10L, 2L)), class = "data.frame", row.names = c(NA,
-8L))
I would like to add another column reporting sum of coupled subsequent rows. Thus the final result would look like this:
I have proceeded like this
Table = as.data.frame(table(data_1$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
CustomerID MarkrtungChannel OrderID
1 A 1
2 B 2
3 A 3
4 B 4
5 C 5
1 C 6
1 A 7
2 C 8
3 B 9
3 B 10
Hi, I want to know which combinations of marketing channels are used by how many customers .
How can I calculate this with R?
E.g. The combination of Marketing channels A and C is used by 1 customer (ID 1)
the combination of Marketing channels C and B is also used by 1 customer (ID 2)
And so on...
and here's a tidyverse way.
library(tidyverse)
data.df%>%
group_by(CustomerID)%>%
summarize(combo=paste0(sort(unique(MarkrtungChannel)),collapse=""))%>%
ungroup()%>%
group_by(combo)%>%
summarize(n.users=n())
counting the number of people using each combo at the end.
You can do it multiple ways. Here is data.table way:
# Here is your data
df<-structure(list(CustomerID = c(1L, 2L, 3L, 4L, 5L, 1L, 1L, 2L,
3L, 3L), MarkrtungChannel = structure(c(1L, 2L, 1L, 2L, 3L, 3L,
1L, 3L, 2L, 2L), .Label = c("A", "B", "C"), class = "factor"),
OrderID = 1:10), .Names = c("CustomerID", "MarkrtungChannel",
"OrderID"), class = "data.frame", row.names = c(NA, -10L))
df[]<-lapply(df[],as.character)
# Here is the combination field
library(data.table)
setDT(df)
df[,Combo:=.(list(unique(MarkrtungChannel))), by=CustomerID]
# Or (to get the combination counts)
df[,list(combo=(list(unique(MarkrtungChannel)))), by=CustomerID][,uniqueN(CustomerID),by=combo]
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))
For a sample dataframe:
df <- structure(list(animal.1 = structure(c(1L, 1L, 2L, 2L, 2L, 4L,
4L, 3L, 1L, 1L), .Label = c("cat", "dog", "horse", "rabbit"), class = "factor"),
animal.2 = structure(c(1L, 2L, 2L, 2L, 4L, 4L, 1L, 1L, 3L,
1L), .Label = c("cat", "dog", "hamster", "rabbit"), class = "factor"),
number = c(5L, 3L, 2L, 5L, 1L, 4L, 6L, 7L, 1L, 11L)), .Names = c("animal.1",
"animal.2","number"), class = "data.frame", row.names = c(NA,
-10L))
... I wish to make a new df with 'animal' duplicates all added together. For example multiple rows with the same animal in columns 1 and 2 will be put together. So for example the dataframe above would read:
cat cat 16
dog dog 7
cat dog 3 etc. etc... (those with different animals would be left as they are). Importantly the sum of 'number' in both dataframes would be the same.
My real df is >400K observations, so anything that anyone could recommend could cope with a large dataset would be great!
Thanks in advance.
One option would be to use data.table. Convert "data.frame" to "data.table" (setDT(), if the "animal.1" rows are equal to "animal.2", then, replace the "number" with sum of "number" after grouping by the two columns, and finally get the unique rows.
library(data.table)
setDT(df)[as.character(animal.1)==as.character(animal.2),
number:=sum(number) ,.(animal.1, animal.2)]
unique(df)
# animal.1 animal.2 number
#1: cat cat 16
#2: cat dog 3
#3: dog dog 7
#4: dog rabbit 1
#5: rabbit rabbit 4
#6: rabbit cat 6
#7: horse cat 7
#8: cat hamster 1
Or an option with dplyr. The approach is similar to data.table. We group by "animal.1", "animal.2", then replace the "number" with sum only when "animal.1" is equal to "animal.2", and get the unique rows
library(dplyr)
df %>%
group_by(animal.1, animal.2) %>%
mutate(number=replace(number,as.character(animal.1)==
as.character(animal.2),
sum(number))) %>%
unique()
I have a dataframe df containing two factor variables (Var and Year) as well as one (in reality several) column with values.
df <- structure(list(Var = structure(c(1L, 1L, 1L, 2L, 2L, 2L, 2L,
3L, 3L, 3L), .Label = c("A", "B", "C"), class = "factor"), Year = structure(c(1L,
2L, 3L, 1L, 2L, 3L, 3L, 1L, 2L, 3L), .Label = c("2000", "2001",
"2002"), class = "factor"), Val = structure(c(1L, 2L, 2L, 4L,
1L, 3L, 3L, 5L, 6L, 6L), .Label = c("2", "3", "4", "5", "8",
"9"), class = "factor")), .Names = c("Var", "Year", "Val"), row.names = c(NA,
-10L), class = "data.frame")
> df
Var Year Val
1 A 2000 2
2 A 2001 3
3 A 2002 3
4 B 2000 5
5 B 2001 2
6 B 2002 4
7 B 2002 4
8 C 2000 8
9 C 2001 9
10 C 2002 9
Now I'd like to find rows with the same value for Val for each Var and Year and only keep one of those. So in this example I would like row 7 to be removed.
I've tried to find a solution with plyr using something like
df_new <- ddply(df, .(Var, Year), summarise, !duplicate(Val))
but obviously that is not a function accepted by ddply.
I found this similar question but the plyr solution by Arun only gives me a dataframe with 0 rows and 0 columns and I do not understand the answer well enough to modify it according to my needs.
Any hints on how to go about that?
Non-duplicates of Val by Var and Year are the same as non-duplicates of Val, Var, and Year. You can specify several columns for duplicated (or the whole data frame).
I think this does what you'd like.
df[!duplicated(df), ]
Or.
df[!duplicated(df[, c("Var", "Year", "Val")]), ]
you can just used the unique() function instead of !duplicate(Val)
df_new <- ddply(df, .(Var, Year), summarise, Val=unique(Val))
# or
df_new <- ddply(df, .(Var, Year), function(x) x[!duplicated(x$Val),])
# or if you only have these 3 columns:
df_new <- ddply(df, .(Var, Year), unique)
# with dplyr
df%.%group_by(Var, Year)%.%filter(!duplicated(Val))
hth
You don't need the plyr package here. If your whole dataset consists of only these 3 columns and you need to remove the duplicates, then you can use,
df_new <- unique(df)
Else, if you need to just pick up the first observation for a group by variable list, then you can use the method suggested by Richard. That's usually how I have been doing it.