Hi all I have a got a 2 datasets below. From these 2 datasets(dataset1 is formed from dataset2. I mean the dataset1 is the count of users from dataset2) can we build the the third datasets(expected output)
dataset1
Apps # user Enteries
A 3
B 4
C 6
dataset2
Apps Users
A X
A Y
A Z
B Y
B Y
B Z
B A
C X
C X
C X
C X
C X
C X
Expected output
Apps Entries X Y Z A
A 3 1 1 1
B 4 2 1 1
C 6 6
We can first count first for Apps and Users, get the data in wide format and join with the table for count of Apps.
library(dplyr)
df %>%
count(Apps, Users) %>%
tidyr::pivot_wider(names_from = Users, values_from = n,
values_fill = list(n = 0)) %>%
left_join(df %>% count(Apps), by = 'Apps')
# Apps X Y Z A n
# <chr> <int> <int> <int> <int> <int>
#1 A 1 1 1 0 3
#2 B 0 2 1 1 4
#3 C 6 0 0 0 6
I showing 0 is no problem and having a different column order you can use table and rowSums to produce the expected output.
x <- table(dataset2)
cbind(Entries=rowSums(x), x)
# Entries A X Y Z
#A 3 0 1 1 1
#B 4 1 0 2 1
#C 6 0 6 0 0
A solution where you need not have to calculate Total separately and do joins...
This solution uses purrr::pmap and dplyr::mutate for dynamically calculating Total.
library(tidyverse) # dplyr, tidyr, purrr
df %>% count(Apps, Users) %>%
pivot_wider(id_cols = Apps, names_from = Users, values_from = n, values_fill = list(n = 0)) %>%
mutate(Total = pmap_int(.l = select_if(., is.numeric),
.f = sum))
which have output what you need
# A tibble: 3 x 6
Apps X Y Z A Total
<chr> <int> <int> <int> <int> <int>
1 A 1 1 1 0 3
2 B 0 2 1 1 4
3 C 6 0 0 0 6
Related
I have a data frame which looks like this:
Value1 = c("1","2","1","3")
Letter = c("A","B","B","A")
A = c("2","2","0","1")
B = c("1","1","1","0")
data <- data.frame(Value1,Letter,A,B)
data
Value1 Letter A B
1 1 A 2 1
2 2 B 2 1
3 1 B 0 1
4 3 A 1 0
I'm trying to add a new column which is the multiplication of column Value1, by column A or B depending on what is in the Letter column. The expected result would be:
Value1 Letter A B Results
1 1 A 2 1 2
2 2 B 2 1 2
3 1 B 0 1 1
4 3 A 1 0 3
I'm trying to use the match() function, but without success.
Thanks!
With base R:
data <- type.convert(data, as.is = TRUE)
data$Results <- ifelse(data$Letter == 'A', data$A * data$Value1, data$B * data$Value1)
Output
Value1 Letter A B Results
1 1 A 2 1 2
2 2 B 2 1 2
3 1 B 0 1 1
4 3 A 1 0 3
Another option would be to pivot to long form, do the calculation, then pivot back to wide format.
library(tidyverse)
data %>%
type.convert(as.is = TRUE) %>%
pivot_longer(c(A, B)) %>%
mutate(Results = ifelse(Letter == name, value * Value1, NA_integer_)) %>%
pivot_wider(names_from = "name", values_from = "value") %>%
group_by(Value1, Letter) %>%
summarise_all(discard, is.na)
Output
Value1 Letter Results A B
<int> <chr> <int> <int> <int>
1 1 A 2 2 1
2 1 B 1 0 1
3 2 B 2 2 1
4 3 A 3 1 0
Use case_when or ifelse
library(dplyr)
data <- data %>%
type.convert(as.is = TRUE) %>%
mutate(Results = case_when(Letter == 'A' ~ A * Value1,
TRUE ~ B * Value1))
-output
data
Value1 Letter A B Results
1 1 A 2 1 2
2 2 B 2 1 2
3 1 B 0 1 1
4 3 A 1 0 3
Or use get with rowwise
data <- data %>%
type.convert(as.is = TRUE) %>%
rowwise %>%
mutate(Result = get(Letter) * Value1) %>%
# or may also use
# mutate(Result = cur_data()[[Letter]] * Value1) %>%
ungroup
-output
data
# A tibble: 4 × 5
Value1 Letter A B Result
<int> <chr> <int> <int> <int>
1 1 A 2 1 2
2 2 B 2 1 2
3 1 B 0 1 1
4 3 A 1 0 3
In base R, we may use row/column indexing as vectorized option
data <- type.convert(data, as.is = TRUE)
nm1 <- unique(data$Letter)
data$Results <-data[nm1][cbind(seq_len(nrow(data)),
match(data$Letter, nm1))] * data$Value1
I have a tibble dt given as follows:
library(tidyverse)
dt <- tibble(x=as.integer(c(0,0,1,0,0,0,1,1,0,1))) %>%
mutate(grp = as.factor(c(rep("A",3), rep("B",4), rep("C",1), rep("D",2))))
dt
As one can observe the rule for grouping is:
starts 0 and ends with 1 (e.g., groups A, B, D) or
it solely contains 1 (e.g., group C)
Problem: Given a tibble with column integer vector x of zeros and 1 that starts with 0 and ends in 1, what is the most efficient way to obtain a grouping using R? (You can use any grouping symbols/factors.)
We can get the cumulative sum of 'x' (assuming it is binary), take the lag add 1 and use that index to replace it with LETTERS (Note that LETTERS was used only as part of matching with the expected output - it can take go up to certain limit)
library(dplyr)
dt %>%
mutate(grp2 = LETTERS[lag(cumsum(x), default = 0)+ 1])
-output
# A tibble: 10 x 3
x grp grp2
<int> <fct> <chr>
1 0 A A
2 0 A A
3 1 A A
4 0 B B
5 0 B B
6 0 B B
7 1 B B
8 1 C C
9 0 D D
10 1 D D
Though the strategy proposed by Akrun is fantastic, yet to show that it can be managed through accumulate also
library(tidyverse)
dt <- tibble(x=as.integer(c(0,0,1,0,0,0,1,1,0,1))) %>%
mutate(grp = as.factor(c(rep("A",3), rep("B",4), rep("C",1), rep("D",2))))
dt %>%
mutate(GRP = accumulate(lag(x, default = 0),.init =1, ~ if(.y != 1) .x else .x+1)[-1])
#> # A tibble: 10 x 3
#> x grp GRP
#> <int> <fct> <dbl>
#> 1 0 A 1
#> 2 0 A 1
#> 3 1 A 1
#> 4 0 B 2
#> 5 0 B 2
#> 6 0 B 2
#> 7 1 B 2
#> 8 1 C 3
#> 9 0 D 4
#> 10 1 D 4
Created on 2021-06-13 by the reprex package (v2.0.0)
I was wondering if there's a more elegant way of taking a dataframe, grouping by x to see how many x's occur in the dataset, then mutating to find the first occurrence of every x (y)
test <- data.frame(x = c("a", "b", "c", "d",
"c", "b", "e", "f", "g"),
y = c(1,1,1,1,2,2,2,2,2))
x y
1 a 1
2 b 1
3 c 1
4 d 1
5 c 2
6 b 2
7 e 2
8 f 2
9 g 2
Current Output
output <- test %>%
group_by(x) %>%
summarise(count = n())
x count
<fct> <int>
1 a 1
2 b 2
3 c 2
4 d 1
5 e 1
6 f 1
7 g 1
Desired Output
x count first_seen
<fct> <int> <dbl>
1 a 1 1
2 b 2 1
3 c 2 1
4 d 1 1
5 e 1 2
6 f 1 2
7 g 1 2
I can filter the test dataframe for the first occurrences then use a left_join but was hoping there's a more elegant solution using mutate?
# filter for first occurrences of y
right <- test %>%
group_by(x) %>%
filter(y == min(y)) %>%
slice(1) %>%
ungroup()
# bind to the output dataframe
left_join(output, right, by = "x")
We can use first after grouping by 'x' to create a new column, use that also in group_by and get the count with n()
library(dplyr)
test %>%
group_by(x) %>%
group_by(first_seen = first(y), add = TRUE) %>%
summarise(count = n())
# A tibble: 7 x 3
# Groups: x [7]
# x first_seen count
# <fct> <dbl> <int>
#1 a 1 1
#2 b 1 2
#3 c 1 2
#4 d 1 1
#5 e 2 1
#6 f 2 1
#7 g 2 1
I have a question. Why not keep it simple? for example
test %>%
group_by(x) %>%
summarise(
count = n(),
first_seen = first(y)
)
#> # A tibble: 7 x 3
#> x count first_seen
#> <chr> <int> <dbl>
#> 1 a 1 1
#> 2 b 2 1
#> 3 c 2 1
#> 4 d 1 1
#> 5 e 1 2
#> 6 f 1 2
#> 7 g 1 2
I have a dataframe df
df <- data.frame(id =c(1,2,1,4,1,5,6),
label=c("a","b", "a", "a","a", "e", "a"),
color = c("g","a","g","g","a","a","a"),
threshold = c(12, 10, 12, 12, 12, 35, 40),
value =c(32.1,0,15.0,10,1,50,45),stringsAsFactors = F
)
Threshold value is based on the label
I should get a table below like this by considering each id,with respective label how many times exceeding its threshold by the value
Color is independent in consideration for calculating the exceed values
I tried like this
final_df <- df %>%
mutate(check = if_else(value > threshold, 1, 0)) %>%
group_by(id, label) %>%
summarise(exceed = sum(check))
But instead of getting with respective id i have got the number of total in exceed
With base R only, use aggregate.
aggregate(seq.int(nrow(df)) ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
# id label seq.int(nrow(df))
#1 1 a 2
#2 4 a 0
#3 6 a 1
#4 2 b 0
#5 5 e 1
In order to match the expected output posted in the question, it will take a little extra work.
exceed <- seq.int(nrow(df))
agg <- aggregate(exceed ~ id + label, df, function(i) sum(df[i, 4] < df[i, 5]))
res <- merge(df[1:3], agg)
unique(res)
# id label color exceed
#1 1 a g 2
#3 1 a a 2
#4 2 b a 0
#5 4 a g 0
#6 5 e a 1
#7 6 a a 1
By a small modification of your code:
df %>%
group_by(id, label) %>%
mutate(check = if_else(value > threshold, 1, 0)) %>%
summarise(exceed = sum(check)) %>%
group_by(id, label)
id label exceed
<dbl> <chr> <dbl>
1 1 a 2
2 2 b 0
3 4 a 0
4 5 e 1
5 6 a 1
To match the expected output more closely:
df %>%
group_by(id, label) %>%
mutate(exceed = sum(if_else(value > threshold, 1, 0))) %>%
group_by(id, label, color) %>%
filter(row_number() == 1)
id label color threshold value exceed
<dbl> <chr> <chr> <dbl> <dbl> <dbl>
1 1 a g 12 32.1 2
2 2 b a 10 0 0
3 4 a g 12 10 0
4 1 a a 12 1 2
5 5 e a 35 50 1
6 6 a a 40 45 1
library(dplyr)
df %>%
group_by(id, label) %>%
mutate(exceed = sum(value > threshold)) %>%
slice(1)
id label color threshold value exceed
<dbl> <chr> <chr> <dbl> <dbl> <int>
1 1 a g 12 32.1 2
2 2 b a 10 0 0
3 4 a g 12 10 0
4 5 e a 35 50 1
5 6 a a 40 45 1
If you like the output to contain a separate row for each combination, of ID, label and color, just add a new group_by before the slice function:
df %>%
group_by(id, label) %>%
mutate(exceed = sum(value > threshold)) %>%
group_by(id, label, color) %>%
slice(1)
id label color threshold value exceed
<dbl> <chr> <chr> <dbl> <dbl> <int>
1 1 a a 12 1 2
2 1 a g 12 32.1 2
3 2 b a 10 0 0
4 4 a g 12 10 0
5 5 e a 35 50 1
6 6 a a 40 45 1
A little change in your code
final_df <- df %>% mutate(check = if_else(value > threshold, 1, 0)) %>% group_by(id, label) %>% filter(check==1)
unique(final_df$id)
We could use table and merge :
table_ <- table(subset(df,value>threshold, c("id","label")))
df2 <- merge(unique(df[c("id","label","color")]),table_,all.x=TRUE)
df2$Freq[is.na(df2$Freq)] <- 0
# id label color Freq
# 1 1 a g 2
# 2 1 a a 2
# 3 2 b a 0
# 4 4 a g 0
# 5 5 e a 1
# 6 6 a a 1
I have a dataframe df with three columns a,b,c.
df <- data.frame(a = c('a','b','c','d','e','f','g','e','f','g'),
b = c('X','Y','Z','X','Y','Z','X','X','Y','Z'),
c = c('cat','dog','cat','dog','cat','cat','dog','cat','cat','dog'))
df
# output
a b c
1 a X cat
2 b Y dog
3 c Z cat
4 d X dog
5 e Y cat
6 f Z cat
7 g X dog
8 e X cat
9 f Y cat
10 g Z dog
I have to group_by using the column b followed by summarise using the column c with counts of available values in it.
df %>% group_by(b) %>%
summarise(nCat = sum(c == 'cat'),
nDog = sum(c == 'dog'))
#output
# A tibble: 3 × 3
b nCat nDog
<fctr> <int> <int>
1 X 2 2
2 Y 2 1
3 Z 2 1
However, before doing the above task, I should remove the rows belonging to a value in a which has more than one value in b.
df %>% group_by(a) %>% summarise(count = n())
#output
# A tibble: 7 × 2
a count
<fctr> <int>
1 a 1
2 b 1
3 c 1
4 d 1
5 e 2
6 f 2
7 g 2
For example, in this dataframe, all the rows having value e(values: Y,X), f(values: Z,Y), g(values: X,Z) in column a.
# Expected output
# A tibble: 3 × 3
b nCat nDog
<fctr> <int> <int>
1 X 1 1
2 Y 0 1
3 Z 1 0
We can use filter with n_distinct to filter the values in 'b' that have only one unique element for each 'a' group, then grouped by 'b', we do the summarise
df %>%
group_by(a) %>%
filter(n_distinct(b)==1) %>%
group_by(b) %>%
summarise(nCat =sum(c=='cat'), nDog = sum(c=='dog'), Total = n())
# A tibble: 3 × 4
# b nCat nDog Total
# <fctr> <int> <int> <int>
#1 X 1 1 2
#2 Y 0 1 1
#3 Z 1 0 1