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I have data that looks like this:
library(dplyr)
Data <- tibble(
ID = c("Code001", "Code001","Code001","Code002","Code002","Code002","Code002","Code002","Code003","Code003","Code003","Code003"),
Value = c(107,107,107,346,346,346,346,346,123,123,123,123))
I need to work out the average value per group per row. However, the value needs to be rounded (so no decimal places) and the group sum needs to equal the group sum of Value.
So solutions like this won't work:
Data %>%
add_count(ID) %>%
group_by(ID) %>%
mutate(Prop_Value_1 = Value/n,
Prop_Value_2 = round(Value/n))
Is there a solution that can produce an output like this:
Data %>%
mutate(Prop_Value = c(35,36,36,69,69,69,69,70,30,31,31,31))
Can use ceiling and then row_number to get there:
Data %>%
group_by(ID) %>%
mutate(count = n(),
ceil_avg = ceiling(Value/count)) %>%
mutate(sum_ceil_avg = sum(ceil_avg),
diff_sum = sum_ceil_avg - Value,
rn = row_number()) %>%
mutate(new_avg = ifelse(rn <= diff_sum,
ceil_avg - 1,
ceil_avg))
# A tibble: 12 × 8
# Groups: ID [3]
ID Value count ceil_avg sum_ceil_avg diff_sum rn new_avg
<chr> <dbl> <int> <dbl> <dbl> <dbl> <int> <dbl>
1 Code001 107 3 36 108 1 1 35
2 Code001 107 3 36 108 1 2 36
3 Code001 107 3 36 108 1 3 36
4 Code002 346 5 70 350 4 1 69
5 Code002 346 5 70 350 4 2 69
6 Code002 346 5 70 350 4 3 69
7 Code002 346 5 70 350 4 4 69
8 Code002 346 5 70 350 4 5 70
9 Code003 123 4 31 124 1 1 30
10 Code003 123 4 31 124 1 2 31
11 Code003 123 4 31 124 1 3 31
12 Code003 123 4 31 124 1 4 31
A first solution is to use integer division:
Data %>%
group_by(ID) %>%
mutate(Prop_Value = ifelse(row_number() <= Value %% n(), Value %/% n() + 1, Value %/% n()))
# A tibble: 12 × 3
# Groups: ID [3]
ID Value Prop_Value
<chr> <dbl> <dbl>
1 Code001 107 36
2 Code001 107 36
3 Code001 107 35
4 Code002 346 70
5 Code002 346 69
6 Code002 346 69
7 Code002 346 69
8 Code002 346 69
9 Code003 123 31
10 Code003 123 31
11 Code003 123 31
12 Code003 123 30
I have a data frame delineated by ownership, private(50) and state(30). Looking to create 5 new rows that are the sum of ownership 50 and ownership 30 as long as they have a matching area value. Desired result is below.
naics <- c(611,611,611,611,611,611,611,611,611,611)
ownership <- c(50,50,50,50,50,30,30,30,30,10)
area <- c(001,003,005,009,011,001,003,005,011,001)
d200201 <- c(14,17,20,23,26,3,5,7,9,100)
d200202 <- c(15,18,21,24,28,9,11,13,15,105)
private <- data.frame(naics,ownership,area,d200201,d200202)
naics ownership area d200201 d200202
611 50 001 17 24
611 50 003 22 29
611 50 005 27 34
611 50 009 23 24 (no sum because no 30 value)
611 50 011 35 43
Is this what you are looking for?
library(dplyr)
private %>%
group_by(naics, area) %>%
summarize(
across(c(d200201, d200202), ~sum(.x[ownership %in% c(30, 50)])),
ownership = 50, .groups = "drop"
)
Output
# A tibble: 5 x 5
naics area d200201 d200202 ownership
<dbl> <dbl> <dbl> <dbl> <dbl>
1 611 1 17 24 50
2 611 3 22 29 50
3 611 5 27 34 50
4 611 9 23 24 50
5 611 11 35 43 50
library(tidyverse)
private %>%
filter(ownership %in% c(50, 30)) %>%
group_by(area) %>%
summarize(across(starts_with("d200"), sum))
#> # A tibble: 5 × 3
#> area d200201 d200202
#> <dbl> <dbl> <dbl>
#> 1 1 17 24
#> 2 3 22 29
#> 3 5 27 34
#> 4 9 23 24
#> 5 11 35 43
Created on 2022-01-08 by the reprex package (v2.0.1)
I have a data frame with lot of company information separated by an id variable. I want to sort one of the variables and repeat it for every id. Let's take this example,
df <- structure(list(id = c(110, 110, 110, 90, 90, 90, 90, 252, 252
), var1 = c(26, 21, 54, 10, 18, 9, 16, 54, 39), var2 = c(234,
12, 43, 32, 21, 19, 16, 34, 44)), .Names = c("id", "var1", "var2"
), row.names = c(NA, -9L), class = "data.frame")
Which looks like this
df
id var1 var2
1 110 26 234
2 110 21 12
3 110 54 43
4 90 10 32
5 90 18 21
6 90 9 19
7 90 16 16
8 252 54 34
9 252 39 44
Now, I want to sort the data frame according to var1 by the vector id. Easiest solution I can think of is using apply function like this,
> apply(df, 2, sort)
id var1 var2
[1,] 90 9 12
[2,] 90 10 16
[3,] 90 16 19
[4,] 90 18 21
[5,] 110 21 32
[6,] 110 26 34
[7,] 110 39 43
[8,] 252 54 44
[9,] 252 54 234
However, this is not the output I am seeking. The correct output should be,
id var1 var2
1 110 21 12
2 110 26 234
3 110 54 43
4 90 9 19
5 90 10 32
6 90 16 16
7 90 18 21
8 252 39 44
9 252 54 34
Group by id and sort by var1 column and keep original id column order.
Any idea how to sort like this?
Note. As mentioned by Moody_Mudskipper, there is no need to use tidyverse and can also be done easily with base R:
df[order(ordered(df$id, unique(df$id)), df$var1), ]
A one-liner tidyverse solution w/o any temp vars:
library(tidyverse)
df %>% arrange(ordered(id, unique(id)), var1)
# id var1 var2
# 1 110 26 234
# 2 110 21 12
# 3 110 54 43
# 4 90 10 32
# 5 90 18 21
# 6 90 9 19
# 7 90 16 16
# 8 252 54 34
# 9 252 39 44
Explanation of why apply(df, 2, sort) does not work
What you were trying to do is to sort each column independently. apply runs over the specified dimension (2 in this case which corresponds to columns) and applies the function (sort in this case).
apply tries to further simplify the results, in this case to a matrix. So you are getting back a matrix (not a data.frame) where each column is sorted independently. For example this row from the apply call:
# [1,] 90 9 12
does not even exist in the original data.frame.
Another base R option using order and match
df[with(df, order(match(id, unique(id)), var1, var2)), ]
# id var1 var2
#2 110 21 12
#1 110 26 234
#3 110 54 43
#6 90 9 19
#4 90 10 32
#7 90 16 16
#5 90 18 21
#9 252 39 44
#8 252 54 34
We can convert the id to factor in order to split while preserving the original order. We can then loop over the list and order, and rbind again, i.e.
df$id <- factor(df$id, levels = unique(df$id))
do.call(rbind, lapply(split(df, df$id), function(i)i[order(i$var1),]))
# id var1 var2
#110.2 110 21 12
#110.1 110 26 234
#110.3 110 54 43
#90.6 90 9 19
#90.4 90 10 32
#90.7 90 16 16
#90.5 90 18 21
#252.9 252 39 44
#252.8 252 54 34
NOTE: You can reset the rownames by rownames(new_df) <- NULL
In base R we could use split<- :
split(df,df$id) <- lapply(split(df,df$id), function(x) x[order(x$var1),] )
or as #Markus suggests :
split(df, df$id) <- by(df, df$id, function(x) x[order(x$var1),])
output in either case :
df
# id var1 var2
# 1 110 21 12
# 2 110 26 234
# 3 110 54 43
# 4 90 9 19
# 5 90 10 32
# 6 90 16 16
# 7 90 18 21
# 8 252 39 44
# 9 252 54 34
With the following tidyverse pipe, the question's output is reproduced.
library(tidyverse)
df %>%
mutate(tmp = cumsum(c(0, diff(id) != 0))) %>%
group_by(id) %>%
arrange(tmp, var1) %>%
select(-tmp)
## A tibble: 9 x 3
## Groups: id [3]
# id var1 var2
# <dbl> <dbl> <dbl>
#1 110 21 12
#2 110 26 234
#3 110 54 43
#4 90 9 19
#5 90 10 32
#6 90 16 16
#7 90 18 21
#8 252 39 44
#9 252 54 34
I'm currently on R trying to create for a DF multiple columns with the sum of previous one. Imagine I got a DF like this:
df=
sep-2016 oct-2016 nov-2016 dec-2016 jan-2017
1 70 153 NA 28 19
2 57 68 73 118 16
3 29 NA 19 32 36
4 177 36 3 54 53
and I want to add at the end the sum of the rows previous of the month that I'm reporting so for October you end up with the sum of sep and oct, and for November you end up with the sum of sep, oct and november and end up with something like this:
df=
sep-2016 oct-2016 nov-2016 dec-2016 jan-2017 status-Oct2016 status-Nov 2016
1 70 153 NA 28 19 223 223
2 57 68 73 118 16 105 198
3 29 NA 19 32 36 29 48
4 177 36 3 54 53 213 93
I want to know a efficient way insted of writing a lots of lines of rowSums() and even if I can get the label on the iteration for each month would be amazing!
Thanks!
We can use lapply to loop through the columns to apply the rowSums.
dat2 <- as.data.frame(lapply(2:ncol(dat), function(i){
rowSums(dat[, 1:i], na.rm = TRUE)
}))
names(dat2) <- paste0("status-", names(dat[, -1]))
dat3 <- cbind(dat, dat2)
dat3
# sep-2016 oct-2016 nov-2016 dec-2016 jan-2017 status-oct-2016 status-nov-2016 status-dec-2016 status-jan-2017
# 1 70 153 NA 28 19 223 223 251 270
# 2 57 68 73 118 16 125 198 316 332
# 3 29 NA 19 32 36 29 48 80 116
# 4 177 36 3 54 53 213 216 270 323
DATA
dat <- read.table(text = " 'sep-2016' 'oct-2016' 'nov-2016' 'dec-2016' 'jan-2017'
1 70 153 NA 28 19
2 57 68 73 118 16
3 29 NA 19 32 36
4 177 36 3 54 53",
header = TRUE, stringsAsFactors = FALSE)
names(dat) <- c("sep-2016", "oct-2016", "nov-2016", "dec-2016", "jan-2017")
Honestly I have no idea why you would want your data in this format, but here is a tidyverse method of accomplishing it. It involves transforming the data to a tidy format before spreading it back out into your wide format. The key thing to note is that in a tidy format, where month is a variable in a single column instead of spread across multiple columns, you can simply use group_by(rowid) and cumsum to calculate all the values you want. The last few lines are constructing the status- column names and spreading the data back out into a wide format.
library(tidyverse)
df <- read_table2(
"sep-2016 oct-2016 nov-2016 dec-2016 jan-2017
70 153 NA 28 19
57 68 73 118 16
29 NA 19 32 36
177 36 3 54 53"
)
df %>%
rowid_to_column() %>%
gather("month", "value", -rowid) %>%
arrange(rowid) %>%
group_by(rowid) %>%
mutate(
value = replace_na(value, 0),
status = cumsum(value)
) %>%
gather("vartype", "number", value, status) %>%
mutate(colname = ifelse(vartype == "value", month, str_c("status-", month))) %>%
select(rowid, number, colname) %>%
spread(colname, number)
#> # A tibble: 4 x 11
#> # Groups: rowid [4]
#> rowid `dec-2016` `jan-2017` `nov-2016` `oct-2016` `sep-2016`
#> <int> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 1 28.0 19.0 0 153 70.0
#> 2 2 118 16.0 73.0 68.0 57.0
#> 3 3 32.0 36.0 19.0 0 29.0
#> 4 4 54.0 53.0 3.00 36.0 177
#> # ... with 5 more variables: `status-dec-2016` <dbl>,
#> # `status-jan-2017` <dbl>, `status-nov-2016` <dbl>,
#> # `status-oct-2016` <dbl>, `status-sep-2016` <dbl>
Created on 2018-02-16 by the reprex package (v0.2.0).
A clean way to do it is by convert your data in a long format.
library(tibble)
library(tidyr)
library(dplyr)
your_data <- tribble(~"sep_2016", ~"oct_2016", ~"nov_2016", ~"dec_2016", ~"jan_2017",
70, 153, NA, 28, 19,
57, 68, 73, 118, 16,
29, NA, 19, 32, 36,
177, 36, 3, 54, 53)
You can change the format of your data.frame with gather from the tidyr package.
your_data_long <- your_data %>%
rowid_to_column() %>%
gather(key = month_year, value = the_value, -rowid)
head(your_data_long)
#> # A tibble: 6 x 3
#> rowid month_year the_value
#> <int> <chr> <dbl>
#> 1 1 sep_2016 70
#> 2 2 sep_2016 57
#> 3 3 sep_2016 29
#> 4 4 sep_2016 177
#> 5 1 oct_2016 153
#> 6 2 oct_2016 68
Once your data.frame is in a long format. You can compute cumulative sum with cumsumand dplyrfunctions mutate and group_by.
result <- your_data_long %>%
group_by(rowid) %>%
mutate(cumulative_value = cumsum(the_value))
result
#> # A tibble: 20 x 4
#> # Groups: rowid [4]
#> rowid month_year the_value cumulative_value
#> <int> <chr> <dbl> <dbl>
#> 1 1 sep_2016 70 70
#> 2 2 sep_2016 57 57
#> 3 3 sep_2016 29 29
#> 4 4 sep_2016 177 177
#> 5 1 oct_2016 153 223
#> 6 2 oct_2016 68 125
#> 7 3 oct_2016 NA NA
#> 8 4 oct_2016 36 213
#> 9 1 nov_2016 NA NA
#> 10 2 nov_2016 73 198
#> 11 3 nov_2016 19 NA
#> 12 4 nov_2016 3 216
#> 13 1 dec_2016 28 NA
#> 14 2 dec_2016 118 316
#> 15 3 dec_2016 32 NA
#> 16 4 dec_2016 54 270
#> 17 1 jan_2017 19 NA
#> 18 2 jan_2017 16 332
#> 19 3 jan_2017 36 NA
#> 20 4 jan_2017 53 323
If you want to retrieve the starting form, you can do it with spread.
My preferred solution would be:
# library(matrixStats)
DF <- as.matrix(df)
DF[is.na(DF)] <- 0
RES <- matrixStats::rowCumsums(DF)
colnames(RES) <- paste0("status-", colnames(DF))
cbind.data.frame(df, RES)
This is closest to what you are looking for with the rowSums.
One option could be using spread and gather function from tidyverse.
Note: The status column has been added even for the 1st month. And the status columns are not in order but values are correct.
The approach is:
# Data
df <- read.table(text = "sep-2016 oct-2016 nov-2016 dec-2016 jan-2017
70 153 NA 28 19
57 68 73 118 16
29 NA 19 32 36
177 36 3 54 53", header = T, stringsAsFactors = F)
library(tidyverse)
# Just add an row number as sl
df <- df %>% mutate(sl = row_number())
#Calculate the cumulative sum after gathering and arranging by date
mod_df <- df %>%
gather(key, value, -sl) %>%
mutate(key = as.Date(paste("01",key, sep="."), format="%d.%b.%Y")) %>%
arrange(sl, key) %>%
group_by(sl) %>%
mutate(status = cumsum(ifelse(is.na(value),0L,value) )) %>%
select(-value) %>%
mutate(key = paste("status",as.character(key, format="%b.%Y"))) %>%
spread(key, status)
# Finally join cumulative calculated sum columns with original df and then
# remove sl column
inner_join(df, mod_df, by = "sl") %>% select(-sl)
# sep.2016 oct.2016 nov.2016 dec.2016 jan.2017 status Dec.2016 status Jan.2017 status Nov.2016 status Oct.2016 status Sep.2016
#1 70 153 NA 28 19 251 270 223 223 70
#2 57 68 73 118 16 316 332 198 125 57
#3 29 NA 19 32 36 80 116 48 29 29
#4 177 36 3 54 53 270 323 216 213 177
Another base solution where we build a matrix accumulating the row sums :
status <- setNames(
as.data.frame(t(apply(dat,1,function(x) Reduce(sum,'[<-'(x,is.na(x),0),accumulate = TRUE)))),
paste0("status-",names(dat)))
status
# status-sep-2016 status-oct-2016 status-nov-2016 status-dec-2016 status-jan-2017
# 1 70 223 223 251 270
# 2 57 125 198 316 332
# 3 29 29 48 80 116
# 4 177 213 216 270 323
Then bind it to your original data if needed :
cbind(dat,status[-1])
Let's say I have got a data.frame like the following:
df = read.table(text = 'A B
11 98
11 87
11 999
11 22
12 34
12 34
12 44
12 98
17 77
17 67
17 87
17 66
33 6
33 45
33 12
33 10', header = TRUE)
I need to group df by col A and select only a given number of rows based on the following vector:
n_rows = c(2, 3, 4, 2)
So that the first group will have only 2 rows (no matter their order), the second group 3 rows, etc...
Here my expected output:
A B
11 98
11 87
12 34
12 34
12 44
17 77
17 67
17 87
17 66
33 6
33 45
I tried to do the trick with dplyr by doing the following:
df %>%
group_by(A) %>%
top_n(n = n_rows, wt =B)
but I got the following error:
Error: n must be a scalar integer
Any suggestion?
thanks
Another base R option,
do.call(rbind, Map(function(x, y) x[seq(y),], split(df, df$A), n_rows))
which gives,
A B
11.1 11 98
11.2 11 87
12.5 12 34
12.6 12 34
12.7 12 44
17.9 17 77
17.10 17 67
17.11 17 87
17.12 17 66
33.13 33 6
33.14 33 45
Here's a possibility, splitting first the data.frame then using map2:
library(dplyr)
library(purr)
df %>% split(.$A) %>%
map2_dfr(n_rows,head)
# A B
# 1 11 98
# 2 11 87
# 3 12 34
# 4 12 34
# 5 12 44
# 6 17 77
# 7 17 67
# 8 17 87
# 9 17 66
# 10 33 6
# 11 33 45
If order doesn't matter you don't need top_n, head works just fine (and faster), else just replace head with top_n.
EDIT:
Here is also a tidy solution, a few characters longer but maybe more satisfying as you don't separate things of the same "kind" but rather work completely inside of the data.frame (same output).
df %>% nest(B) %>%
mutate(data = map2(data,n_rows,head)) %>%
unnest
In base R, you can do something like:
df2 <- data.frame()
for (i in seq_along(unique(df$A))) {
df2 <- rbind(df2, df[df$A == unique(df$A)[i], ][1:n_rows[i], ])
}
> df2
A B
1 11 98
2 11 87
5 12 34
6 12 34
7 12 44
9 17 77
10 17 67
11 17 87
12 17 66
13 33 6
14 33 45
Here is an option with top_n
library(tidyverse)
df %>%
split(., .$A) %>%
map2_df(., n_rows, ~ top_n(., .y, wt = .$B))
If we are not looking for top_n, then another option is slice
df %>%
group_by(A) %>%
nest(B) %>%
mutate(newcol = map2(data, n_rows, ~ .x %>% slice(seq(.y)))) %>%
select(-data) %>%
unnest