I have a dataframe ("df") with a number of columns that I would like to estimate the weighted means of, weighting by population (df$Population), and grouping by commuting zone (df$cz).
This is the list of columns I would like to estimate the weighted means of:
vlist = c("Public_Welf_Total_Exp", "Welf_Cash_Total_Exp", "Welf_Cash_Cash_Assist", "Welf_Ins_Total_Exp","Total_Educ_Direct_Exp", "Higher_Ed_Total_Exp", "Welf_NEC_Cap_Outlay","Welf_NEC_Direct_Expend", "Welf_NEC_Total_Expend", "Total_Educ_Assist___Sub", "Health_Total_Expend", "Total_Hospital_Total_Exp", "Welf_Vend_Pmts_Medical","Hosp_Other_Total_Exp","Unemp_Comp_Total_Exp", "Unemp_Comp_Cash___Sec", "Total_Unemp_Rev", "Hous___Com_Total_Exp", "Hous___Com_Construct")
This is the code I have been using:
df = df %>% group_by(cz) %>% mutate_at(vlist, weighted.mean(., df$Population))
I have also tried:
df = df %>% group_by(cz) %>% mutate_at(vlist, function(x) weighted.mean(x, df$Population))
As well as tested the following code on only 2 columns:
df = df %>% group_by(cz) %>% mutate_at(vars(Public_Welf_Total_Exp, Welf_Cash_Total_Exp), weighted.mean(., df$Population))
However, everything I have tried gives me the following error, even though there are no NAs in any of my variables:
Error in weighted.mean.default(., df$Population) :
'x' and 'w' must have the same length
I understand that I could do the following estimation using lapply, but I don't know how to group by another variable using lapply. I would appreciate any suggestions!
There is a lot to unpack here...
Probably you mean summarise instead of mutate, because with mutate you would just replicate your result for each row.
mutate_at and summarise_at are subseeded and you should use across instead.
the reason why your code wasn't working was because you did not write your function as a formula (you did not add ~ at the beginning), also you were using df$Population instead of Population. When you write Population, summarise knows you're talking about the column Population which, at that point, is grouped like the rest of the dataframe. When you use df$Population you are calling the column of the original dataframe without grouping. Not only it is wrong, but you would also get an error because the length of the variable you are trying to average and the lengths of the weights provided by df$Population would not correspond.
Here is how you could do it:
library(dplyr)
df %>%
group_by(cz) %>%
summarise(across(vlist, weighted.mean, Population),
.groups = "drop")
If you really need to use summarise_at (and probably you are using an old version of dplyr [lower than 1.0.0]), then you could do:
df %>%
group_by(cz) %>%
summarise_at(vlist, ~weighted.mean(., Population)) %>%
ungroup()
I considered df and vlist like the following:
vlist <- c("Public_Welf_Total_Exp", "Welf_Cash_Total_Exp", "Welf_Cash_Cash_Assist", "Welf_Ins_Total_Exp","Total_Educ_Direct_Exp", "Higher_Ed_Total_Exp", "Welf_NEC_Cap_Outlay","Welf_NEC_Direct_Expend", "Welf_NEC_Total_Expend", "Total_Educ_Assist___Sub", "Health_Total_Expend", "Total_Hospital_Total_Exp", "Welf_Vend_Pmts_Medical","Hosp_Other_Total_Exp","Unemp_Comp_Total_Exp", "Unemp_Comp_Cash___Sec", "Total_Unemp_Rev", "Hous___Com_Total_Exp", "Hous___Com_Construct")
df <- as.data.frame(matrix(rnorm(length(vlist) * 100), ncol = length(vlist)))
names(df) <- vlist
df$cz <- rep(letters[1:10], each = 10)
df$Population <- runif(100)
Related
I've got a huge df that include the following:
subsetdf <- data_frame(Id=c(1:6),TicketNo=c(15,16,15,17,17,17))
I want to add a column, GroupSize, that tells for each Id how many other Ids share the same TicketNo value. In other words, I want output like this:
TheDream <- data_frame(Id=c(1:6),TicketNo=c(15,16,15,17,17,17),GroupSize=c(2,1,2,3,3,3)
I've unsuccessfully tried:
subsetdf <- subsetdf %>%
group_by(TicketNo) %>%
add_count(name = "GroupSize")
I'd like to use mutate() but I can't seem to get it right.
Edit
With the GroupSize column now added, I want to add a final column that looks at the values in two other columns and returns the value of whichever is higher. So I've got:
df <- data_frame(Id=c(1:6),TicketNo=c(15,16,15,17,17,17),GroupSize=c(2,1,2,3,3,3),FamilySize=c(2,2,1,1,4,4)
And I want:
df <- data_frame(Id=c(1:6),TicketNo=c(15,16,15,17,17,17),GroupSize=c(2,1,2,3,3,3),FamilySize=c(2,2,1,1,4,4),FinalSize=c(2,2,2,3,4,4)
I've unsuccessfully tried:
df <- df %>% pmax(df$GroupSize, df$FamilySize) %>% dplyr::mutate(FinalSize = n())
That attempt earns me the error: Error: ! Subscript iis a matrix, the datavalue` must have size 1.
Backtrace:
... %>% dplyr::mutate(Groupsize = n())
base::pmax(., train_data$Family_size, train_data$PartySize)
tibble:::[<-.tbl_df(*tmp*, change, value = <int>)
tibble:::tbl_subassign_matrix(x, j, value, j_arg, substitute(value))`
If we need to use mutate use n() to get the group size. Also, make sure that the mutate is from dplyr (as there is also a plyr::mutate which could mask the function if it is loaded later)
library(dplyr)
subsetdf %>%
group_by(TicketNo) %>%
dplyr::mutate(GroupSize = n())
I would like to create lagged values for multiple columns in R.
First, I used a function to create lead/lag like this:
mleadlag <- function(x, n, ts_id) {
pos <- match(as.numeric(ts_id) + n, as.numeric(ts_id))
x[pos]
}
Second, I would like to apply this function for several columns in R. firm.characteristics is list of columns I would like to compute lagged values.
library(dplyr)
firm.characteristics <- colnames(df)[4:6]
for(i in 1:length(firm.characteristics)){
df <- df %>%
group_by(company) %>%
mutate(!!paste0("lag_", i) := mleadlag(df[[i]] ,-1, fye)) %>%
ungroup()
}
However, I didn't get the correct values. The output for all companies in year t is the last row in year t-1. It didn't group by the company any compute the lagged values.
Can anyone help me which is wrong in the loop? Or what should I do to get the correct lagged values?
Thank you so much for your help.
Reproducible sample could be like this:
set.seed(42) ## for sake of reproducibility
n <- 6
dat <- data.frame(company=1:n,
fye=2009,
x=rnorm(n),
y=rnorm(n),
z=rnorm(n),
k=rnorm(n),
m=rnorm(n))
dat2 <- data.frame(company=1:n,
fye=2010,
x=rnorm(n),
y=rnorm(n),
z=rnorm(n),
k=rnorm(n),
m=rnorm(n))
dat3 <- data.frame(company=1:n,
fye=2011,
x=rnorm(n),
y=rnorm(n),
z=rnorm(n),
k=rnorm(n),
m=rnorm(n))
df <- rbind(dat,dat2,dat3)
I would try to stay away from loops in the tidyverse. Many of the tidyverse applications that would traditionally require loops already exist and are very fast, which creates more efficient and intuitive code (the latter being my opinion). This is a great use case for dplyr's across() functionality. I first changed the df to a tibble.
df %>%
as_tibble() %>%
group_by(company) %>%
mutate(
across(firm.characteristics, ~lag(., 1L))
) %>%
ungroup()
This generates the required lagged values. For more information see dplyr's across documentation.
I tried to do a t-test comparing values between time1/2/3.. and threshold.
here is my data frame:
time.df1<-data.frame("condition" =c("A","B","C","A","C","B"),
"time1" = c(1,3,2,6,2,3) ,
"time2" = c(1,1,2,8,2,9) ,
"time3" = c(-2,12,4,1,0,6),
"time4" = c(-8,3,2,1,9,6),
"threshold" = c(-2,3,8,1,9,-3))
and I tried to compare each two values by:
time.df1%>%
select_if(is.numeric) %>%
purrr::map_df(~ broom::tidy(t.test(. ~ threshold)))
However, I got this error message
Error in eval(predvars, data, env) : object 'threshold' not found
So, I tried another way (maybe it is wrong)
time.df2<-time.df1%>%gather(TF,value,time1:time4)
time.df2%>% group_by(condition) %>% do(tidy(t.test(value~TF, data=.)))
sadly, I got this error. Even I limited the condition to only two levels (A,B)
Error in t.test.formula(value ~ TF, data = .) : grouping factor must have exactly 2 levels
I wish to loop t-test over each time column to threshold column per condition, then using broom::tidy to get the results in tidy format. My approaches apparently aren't working, any advice is much appreciated to improve my codes.
An alternative route would be to define a function with the required options for t.test() up front, then create data frames for each pair of variables (i.e. each combination of 'time*' and 'threshold') and nesting them into list columns and use map() combined with relevant functions from 'broom' to simplify the outputs.
library(tidyverse)
library(broom)
ttestfn <- function(data, ...){
# amend this function to include required options for t.test
res = t.test(data$resp, data$threshold)
return(res)
}
df2 <-
time.df1 %>%
gather(time, "resp", - threshold, -condition) %>%
group_by(time) %>%
nest() %>%
mutate(ttests = map(data, ttestfn),
glances = map(ttests, glance))
# df2 has data frames, t-test objects and glance summaries
# as separate list columns
Now it's easy to query this object to extract what you want
df2 %>%
unnest(glances, .drop=TRUE)
However, it's unclear to me what you want to do with 'condition', so I'm wondering if it is more straightforward to reframe the question in terms of a GLM (as camille suggested in the comments: ANOVA is part of the GLM family).
Reshape the data, define 'threshold' as the reference level of the 'time' factor and the default 'treatment' contrasts used by R will compare each time to 'threshold':
time.df2 <-
time.df1 %>%
gather(key = "time", value = "resp", -condition) %>%
mutate(time = fct_relevel(time, "threshold")) # define 'threshold' as baseline
fit.aov <- aov(resp ~ condition * time, data = time.df2)
summary(fit.aov)
summary.lm(fit.aov) # coefficients and p-values
Of course this assumes that all subjects are independent (i.e. there are no repeated measures). If not, then you'll need to move on to more complicated procedures. Anyway, moving to appropriate GLMs for the study design should help minimise the pitfalls of doing multiple t-tests on the same data set.
We could remove the threshold from the select and then reintroduce it by creating a data.frame which would go into the formula object of t.test
library(tidyverse)
time1.df %>%
select_if(is.numeric) %>%
select(-threshold) %>%
map_df(~ data.frame(time = .x, time1.df['threshold']) %>%
broom::tidy(t.test(. ~ threshold)))
I have a large data frame with on every rows enough data to calculate a correlation using specific columns of this data frame and add a new column containing the correlations calculated.
Here is a summary of what I would like to do (this one using dplyr):
example_data %>%
mutate(pearsoncor = cor(x = X001_F5_000_A:X030_F5_480_C, y = X031_H5_000_A:X060_H5_480_C))
Obviously it is not working this way as I get only NA's in the pearsoncor column, does anyone has a suggestion? Is there an easy way to do this?
Best,
Example data frame
With tidyr, you can gather separately all x- and y-variables, you'd like to compare. You get a tibble containing the correlation coefficients and their p-values for every combination you provided.
library(dplyr)
library(tidyr)
example_data %>%
gather(x_var, x_val, X001_F5_000_A:X030_F5_480_C) %>%
gather(y_var, y_val, X031_H5_000_A:X060_H5_480_C) %>%
group_by(x_var, y_var) %>%
summarise(cor_coef = cor.test(x_val, y_val)$estimate,
p_val = cor.test(x_val, y_val)$p.value)
edit, update some years later:
library(tidyr)
library(purrr)
library(broom)
library(dplyr)
longley %>%
pivot_longer(GNP.deflator:Armed.Forces, names_to="x_var", values_to="x_val") %>%
pivot_longer(Population:Employed, names_to="y_var", values_to="y_val") %>%
nest(data=c(x_val, y_val)) %>%
mutate(cor_test = map(data, ~cor.test(.x$x_val, .x$y_val)),
tidied = map(cor_test, tidy)) %>%
unnest(tidied)
Here is a solution using the reshape2 package to melt() the data frame into long form so that each value has its own row. The original wide-form data has 60 values per row for each of the 6 genes, while the melted long-form data frame has 360 rows, one for each value. Then we can easily use summarize() from dplyr to calculate the correlations without loops.
library(reshape2)
library(dplyr)
names1 <- names(example_data)[4:33]
names2 <- names(example_data)[34:63]
example_data_longform <- melt(example_data, id.vars = c('Gene','clusterFR','clusterHR'))
example_data_longform %>%
group_by(Gene, clusterFR, clusterHR) %>%
summarize(pearsoncor = cor(x = value[variable %in% names1],
y = value[variable %in% names2]))
You could also generate more detailed results, as in Eudald's answer, using do():
detailed_r <- example_data_longform %>%
group_by(Gene, clusterFR, clusterHR) %>%
do(cor = cor.test(x = .$value[.$variable %in% names1],
y = .$value[.$variable %in% names2]))
This outputs a tibble with the cor column being a list with the results of cor.test() for each gene. We can use lapply() to extract output from the list.
lapply(detailed_r$cor, function(x) c(x$estimate, x$p.value))
I had the same problem a few days back, and I know loops are not optimal in R but that's the only thing I could think of:
df$r = rep(0,nrow(df))
df$cor_p = rep(0,nrow(df))
for (i in 1:nrow(df)){
ct = cor.test(as.numeric(df[i,cols_A]),as.numeric(df[i,cols_B]))
df$r[i] = ct$estimate
df$cor_p[i] = ct$p.value
}
I am currently learning purrr in R. I have code which does the following
Uses the pysch package in r to get the mean, SD, range etc from a list of questions
Returns those statistics in a single data-frame where the list item is added to the table as a column. In the case below its schools.
Below is an example where I'm about 90% there i think. All i want to do is add the names of the schools to the dataframe as a column so as to be able to chart them afterwards. Can anyone help? The method below loses the names as soon as the bind_rows() command is run
library(lavaan)
library(tidyverse)
# function pulls the mean, sd, range, kurtosis and skew
get_stats <- function(x){
row_names <- rownames(x)
mydf_temp <- x %>%
dplyr::select(mean, sd, range, kurtosis, skew) %>%
mutate_if(is.numeric, round, digits=2) %>%
filter(complete.cases(.))
mydf_temp
}
# Generate the data for the reproducible example
mydf <- HolzingerSwineford1939 %>%
select(school, starts_with("x")) %>%
psych::describeBy(., group=.$school, digits = 2)
# Gets the summary statistics per school
stats_summ <- mydf %>%
map(get_stats) %>%
bind_rows()
We can use the .id argument from bind_rows
mydf %>%
map(get_stats) %>%
bind_rows(., .id = 'group')
Using a reproducible example with iris dataset
mydf <- iris %>%
psych::describeBy(., group=.$Species, digits = 2)