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I have a list of dataframes. It looks something like this:
df1 <- data.frame(Var1 = c(1, 7, 9, 4, 2),
Var2 = c(7, 2, 4, 4, 3),
Var3 = c(3, 6, 2, 0, 8))
df2 <- data.frame(Var1 = c(5, 6, 2, 2, 1),
Var2 = c(8, 6, 6, 7, 4),
Var3 = c(9, 0, 1, 3, 4))
df3.wxyz <- data.frame(Var1 = c("w", "x", "y", "z", 3, 7, 3, 6, 6),
Var2 = c(NA, NA, NA, NA, 7, 5, 8, 0, 2),
Var3 = c(NA, NA, NA, NA, 3, 3, 4, 1, 9))
df4 <- data.frame(Var1 = c(2, 7, 2, 4, 8),
Var2 = c(8, 3, 1, 7, 3),
Var3 = c(9, 1, 1, 6, 5))
df5.wxyz <- data.frame(Var1 = c("w", "x", "y", "z", 2, 7, 3, 1, 6),
Var2 = c(NA, NA, NA, NA, 7, 4, 8, 1, 9),
Var3 = c(NA, NA, NA, NA, 8, 0, 4, 1, 2))
df.list <- list(df1, df2, df3.wxyz, df4, df5.wxyz)
names(df.list) <- c("df1", "df2", "df3.wxyz", "df4", "df5.wxyz")
I would like to remove the first 4 rows of df3.wxyz and df5.wxyz from the list of dataframes as those contain information that I do not need. What I've tried is the following code, but instead of only removing the first 4 rows in df3.wxyz and df5.wxyz, it is removing the first 4 rows from every dataframe in my list. I'm not sure what the issue is.
df.list <- lapply(df.list, function(i){
ifelse(grepl("wxyz", names(df.list)), i <- i[-c(1:4), ], df.list)
i
})
This is what I would like to achieve:
df1 <- data.frame(Var1 = c(1, 7, 9, 4, 2),
Var2 = c(7, 2, 4, 4, 3),
Var3 = c(3, 6, 2, 0, 8))
df2 <- data.frame(Var1 = c(5, 6, 2, 2, 1),
Var2 = c(8, 6, 6, 7, 4),
Var3 = c(9, 0, 1, 3, 4))
df3.wxyz <- data.frame(Var1 = c(3, 7, 3, 6, 6),
Var2 = c(7, 5, 8, 0, 2),
Var3 = c(3, 3, 4, 1, 9))
df4 <- data.frame(Var1 = c(2, 7, 2, 4, 8),
Var2 = c(8, 3, 1, 7, 3),
Var3 = c(9, 1, 1, 6, 5))
df5.wxyz <- data.frame(Var1 = c(2, 7, 3, 1, 6),
Var2 = c(7, 4, 8, 1, 9),
Var3 = c(8, 0, 4, 1, 2))
df.list <- list(df1, df2, df3.wxyz, df4, df5.wxyz)
names(df.list) <- c("df1", "df2", "df3.wxyz", "df4", "df5.wxyz")
You can try,
df.list[grepl('wxyz', names(df.list))] <- lapply(df.list[grepl('wxyz', names(df.list))], na.omit)
You can try na.omit like below
> Map(na.omit,df.list)
$df1
Var1 Var2 Var3
1 1 7 3
2 7 2 6
3 9 4 2
4 4 4 0
5 2 3 8
$df2
Var1 Var2 Var3
1 5 8 9
2 6 6 0
3 2 6 1
4 2 7 3
5 1 4 4
$df3.wxyz
Var1 Var2 Var3
5 3 7 3
6 7 5 3
7 3 8 4
8 6 0 1
9 6 2 9
$df4
Var1 Var2 Var3
1 2 8 9
2 7 3 1
3 2 1 1
4 4 7 6
5 8 3 5
$df5.wxyz
Var1 Var2 Var3
5 2 7 8
6 7 4 0
7 3 8 4
8 1 1 1
9 6 9 2
I have a list named df which contains three iterations with two years of projection.
What I want is: weighting the variable "district" just for year 2 in each iteration and finally I want to have mean of each weighted district for all three iterations. Note that each year has a variable named "weight" that weighting should be based on this variable.
iteration1 <- list(year1 = data.frame(age = c(10, 11, 12, 13),
district = c(1, 2, 3, 4),
gender = c(1, 2, 2, 1),
weight = c(12.2, 11.3, 11.2, 10.1)),
year2 = data.frame(age = c(10, 11, 12, 13, 10, 10),
district = c(1, 2, 3, 4, 2, 1),
gender = c(1, 2, 2, 1, 1, 1),
weight = c(12.2, 11.3, 11.2, 10.1, 12.2, 13.1)))
iteration2 <- list(year1 = data.frame(age = c(10, 11, 12, 13),
district = c(1, 2, 3, 4),
gender = c(2, 2, 1, 1),
weight = c(12.2, 11.3, 11.2, 10.1)),
year2 = data.frame(age = c(10, 11, 12, 13, 13, 13, 12),
district = c(1, 2, 3, 4, 1, 3, 3),
gender = c(2, 2, 1, 1, 2, 2, 2),
weight = c(12.2, 11.3, 11.2, 10.1, 10.9, 11.9, 15.1)))
iteration3 <- list(year1 = data.frame(age = c(10, 11, 12, 13),
district = c(1, 2, 3, 4),
gender = c(2, 2, 1, 1),
weight = c(12.2, 11.3, 11.2, 10.1)),
year2 = data.frame(age = c(10, 11, 12, 13, 10, 10, 11, 12),
district = c(1, 2, 3, 4, 4, 3, 2, 2),
gender = c(2, 2, 1, 1, 2, 2, 1, 2),
weight = c(12.2, 11.3, 11.2, 10.1, 13.5, 12.8, 13.9, 14.9)))
df <- list(iteration1 = iteration1, iteration2 = iteration2, iteration3 = iteration3)
Expected output:
district mean of each district for all three iterations
1 20.2
2 24.96
3 24.46
4 14.6
for calculating my expected output I have followed two steps. in first stepŘ I have weighted year 2 in each iteration by wtd.table(df$iteration1$year2$district,weights=df$iteration1$year2$weight) . I repeated this code for three times (because I have three iterations). here is my output:
1 2 3 4
25.3 23.5 11.2 10.1
1 2 3 4
23.1 11.3 38.2 10.1
1 2 3 4
12.2 40.1 24.0 23.6
in second step, I calculate mean of each district for three iterations manually: mean(25.3,23.1,12.2)
data.table approach
library(data.table)
library(questionr)
ans <- rbindlist(
lapply(df, function(x)
as.data.table(
questionr::wtd.table(x[["year2"]]$district,
weights = x[["year2"]]$weight))),
use.names = TRUE, fill = TRUE)
# Summarise
ans[, .(weight = mean(N, na.rm = TRUE)), by = .(district = V1)]
# district weight
# 1: 1 20.20000
# 2: 2 24.96667
# 3: 3 24.46667
# 4: 4 14.60000
Version 2
With updated columns based on TS's comment below
ans <- rbindlist(
lapply(df, function(x)
as.data.table(
questionr::wtd.table(x = x[["year2"]]$district,
y = x[["year2"]]$gender,
weights = x[["year2"]]$weight) ) ),
use.names = TRUE, fill = TRUE )
# Summarise
ans[, .(n = .N,
mean = mean(N, na.rm = TRUE),
sd = sd(N, na.rm = TRUE)),
by = .(district = V1, gender = V2)]
# district gender n mean sd
# 1: 1 1 3 8.433333 14.606962
# 2: 2 1 3 8.700000 7.582216
# 3: 3 1 3 7.466667 6.466323
# 4: 4 1 3 10.100000 0.000000
# 5: 1 2 3 11.766667 11.556095
# 6: 2 2 3 16.266667 8.602519
# 7: 3 2 3 17.000000 8.697126
# 8: 4 2 3 4.500000 7.794229
Combine the list of dataframes into one and calculate average weight using questionr::wtd.table for each district and iteration in year2. Finally, get aggregated mean for each district.
Using tidyverse you can do -
library(dplyr)
library(purrr)
map_df(df, ~bind_rows(.x, .id = 'year'), .id = 'iter') %>%
filter(year == 'year2') %>%
group_by(district, iter) %>%
summarise(result = questionr::wtd.table(district,weights=weight)) %>%
summarise(result = mean(result))
# district result
# <dbl> <dbl>
#1 1 20.2
#2 2 25.0
#3 3 24.5
#4 4 14.6
I am trying to sort my data in descending or ascending order regardless of the data in the rows. I made a dummy example below:
A <- c(9,9,5,4,6,3,2,NA)
B <- c(9,5,3,4,1,4,NA,NA)
C <- c(1,4,5,6,7,4,2,4)
base <- data.frame(A,B,C)
df <- base
df$A <- sort(df$A,na.last = T)
df$B <- sort(df$B,na.last = T)
df$C <- sort(df$C)
We get this
structure(list(A = c(2, 3, 3, 4, 4, 4, 5, 5, 6, 9, 9, NA), B = c(1,
2, 3, 4, 4, 4, 5, 5, 9, 10, NA, NA), C = c(1, 2, 3, 4, 4, 4,
5, 5, 6, 7, 8, 8)), row.names = c(NA, -12L), class = "data.frame")
I want to get something similar to df but my data have hundreds of columns, is there an easier way to do it?
I tried arrange_all() but the result is not what i want.
library(tidyverse)
test <- base%>%
arrange_all()
Obtaining this:
structure(list(A = c(2, 3, 3, 4, 4, 4, 5, 5, 6, 9, 9, NA), B = c(NA,
2, 4, 4, 5, 10, 3, 4, 1, 5, 9, NA), C = c(2, 3, 4, 6, 8, 5, 5,
8, 7, 4, 1, 4)), class = "data.frame", row.names = c(NA, -12L
))
You can sort each column individually :
library(dplyr)
base %>% mutate(across(.fns = sort, na.last = TRUE))
# A B C
#1 2 1 1
#2 3 3 2
#3 4 4 4
#4 5 4 4
#5 6 5 4
#6 9 9 5
#7 9 NA 6
#8 NA NA 7
Or in base R :
base[] <- lapply(base, sort, na.last = TRUE)
I have a cvs file that has the following structure (minimum example):
ID Variable Vector
1 a [0,0,0]
2 a [1,2,3]
1 a [1,1,2]
2 a [1,2,3]
1 b [0,0,0]
2 b [1,1,1]
1 b [0,0,1]
2 b [3,5,7]
I would like to calculate the mean vector for each combination of parameters (in this case, ID and Variable). That is, I want to obtain a dataframe like the following one:
ID Variable Vector
1 a [0.5,0.5,1]
2 a [1,2,3]
1 b [0,0,0.5]
2 b [2,3,4]
I have generated this csv file with Python, that's why I have that structure with brackets. But I do not know how to start to do this using R. It doesn't seem to be a common data structure.
Update:
Vector variable structure (obtained from dput(head(data, 8))
Vector = c("[3, 16, 14, 5, 6, 13, 17, 7, 13, 6]",
"[7, 12, 6, 10, 6, 5, 16, 9, 19, 10]", "[4, 13, 4, 11, 6, 15, 17, 10, 12, 8]",
"[18, 11, 16, 8, 10, 10, 7, 4, 9, 7]", "[9, 9, 10, 17, 8, 13, 3, 13, 8, 10]",
"[17, 12, 7, 13, 6, 13, 8, 9, 5, 10]", "[9, 6, 14, 10, 8, 4, 8, 14, 15, 12]",
"[7, 13, 8, 10, 16, 8, 13, 13, 8, 4]")), row.names = c(NA, 8L
), class = "data.frame")
Assuming the 'Vector' column is a list, after grouping by 'ID', 'Variable', we reduce the 'Vector' by adding (+) the corresponding elements together and then divide by the total number of elements (n()) in that group
library(dplyr)
library(purrr)
out <- df1 %>%
group_by(ID, Variable) %>%
summarise(Vector = list(reduce(Vector, `+`)/n()), .groups = 'drop')
-output
out
# A tibble: 4 x 3
# ID Variable Vector
# <dbl> <chr> <list>
#1 1 a <dbl [3]>
#2 1 b <dbl [3]>
#3 2 a <dbl [3]>
#4 2 b <dbl [3]>
out$Vector
#[[1]]
#[1] 0.5 0.5 1.0
#[[2]]
#[1] 0.0 0.0 0.5
#[[3]]
#[1] 1 2 3
#[[4]]
#[1] 2 3 4
If the column 'Vector' is a character string, an option is to extract the numeric part into a list
library(stringr)
out <- df1 %>%
group_by(ID, Variable) %>%
summarise(Vector = list((str_extract_all(Vector, "\\d+") %>%
map(as.numeric) %>% reduce(`+`))/n()), .groups = 'drop')
data
df1 <- structure(list(ID = c(1, 2, 1, 2, 1, 2, 1, 2), Variable = c("a",
"a", "a", "a", "b", "b", "b", "b"), Vector = structure(list(c(0,
0, 0), c(1, 2, 3), c(1, 1, 2), c(1, 2, 3), c(0, 0, 0), c(1, 1,
1), c(0, 0, 1), c(3, 5, 7)), class = "AsIs")), class = "data.frame",
row.names = c(NA,
-8L))
This question already has answers here:
How collect additional row data on binned data in R
(1 answer)
Group value in range r
(3 answers)
Closed 3 years ago.
I am doing a statistic analysis in a big data frame (more than 48.000.000 rows) in r. Here is an exemple of the data:
structure(list(herd = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3), cows = c(1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 2, 3, 4,
5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11, 12, 13, 14, 15, 16), `date` = c("11/03/2013",
"12/03/2013", "13/03/2013", "14/03/2013", "15/03/2013", "16/03/2013",
"13/05/2012", "14/05/2012", "15/05/2012", "16/05/2012", "17/05/2012",
"18/05/2012", "10/07/2016", "11/07/2016", "12/07/2016", "13/07/2016",
"11/03/2013", "12/03/2013", "13/03/2013", "14/03/2013", "15/03/2013",
"16/03/2013", "13/05/2012", "14/05/2012", "15/05/2012", "16/05/2012",
"17/05/2012", "18/05/2012", "10/07/2016", "11/07/2016", "12/07/2016",
"13/07/2016", "11/03/2013", "12/03/2013", "13/03/2013", "14/03/2013",
"15/03/2013", "16/03/2013", "13/05/2012", "14/05/2012", "15/05/2012",
"16/05/2012", "17/05/2012", "18/05/2012", "10/07/2016", "11/07/2016",
"12/07/2016", "13/07/2016"), glicose = c(240666, 23457789, 45688688,
679, 76564, 6574553, 78654, 546432, 76455643, 6876, 7645432,
876875, 98654, 453437, 98676, 9887554, 76543, 9775643, 986545,
240666, 23457789, 45688688, 679, 76564, 6574553, 78654, 546432,
76455643, 6876, 7645432, 876875, 98654, 453437, 98676, 9887554,
76543, 9775643, 986545, 240666, 23457789, 45688688, 679, 76564,
6574553, 78654, 546432, 76455643, 6876)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -48L))
I need to identify how many cows are in the following category of glicose by herd and by date:
<=100000
100000 and <=150000
150000 and <=200000
200000 and <=250000
250000 and <=400000
>400000
I tried to use the functions filter() and select() but could not categorize the variable like that.
I tried either to make a vector for each category but it did not work:
ht <- df %>% group_by(herd, date) %>%
filter(glicose < 100000)
Actually I do not have a clue of how I could do this. Please help!
I expect to get the number of cows in each category of each herd based on each date in a table like this:
Calling your data df,
df %>%
mutate(glicose_group = cut(glicose, breaks = c(0, seq(1e5, 2.5e5, by = 0.5e5), 4e5, Inf)),
date = as.Date(date, format = "%d/%m/%Y")) %>%
group_by(herd, date, glicose_group) %>%
count
# # A tibble: 48 x 4
# # Groups: herd, date, glicose_group [48]
# herd date glicose_group n
# <dbl> <date> <fct> <int>
# 1 1 2012-05-13 (0,1e+05] 1
# 2 1 2012-05-14 (4e+05,Inf] 1
# 3 1 2012-05-15 (4e+05,Inf] 1
# 4 1 2012-05-16 (0,1e+05] 1
# 5 1 2012-05-17 (4e+05,Inf] 1
# 6 1 2012-05-18 (4e+05,Inf] 1
# 7 1 2013-03-11 (2e+05,2.5e+05] 1
# 8 1 2013-03-12 (4e+05,Inf] 1
# 9 1 2013-03-13 (4e+05,Inf] 1
# 10 1 2013-03-14 (0,1e+05] 1
# # ... with 38 more rows
I also threw in a conversion to Date class, which is probably a good idea.