relatively new to r, I have a list of words I want to run through the gtrendsr function to look at the google search hits, and then create a tibble with dates as index and relevant hits for each word as columns, I'm struggling to do this using the map functions in purr,
I started off trying to use a for loop but I've been told to try and use map in the tidyverse package instead, this is what I had so far:
library(gtrendsr)
words = c('cruise', 'plane', 'car')
for (i in words) {
rel_word_data = gtrends(i,geo= '', time = 'today 12-m')
iot <- data.frame()
iot[i] <- rel_word_data$interest_over_time$hits
}
I need to have the gtrends function take one word at a time, otherwise it will give a value for hits which is a adjusted for the popularity of the other words. so basically, I need the gtrends function to run the first word in the list, obtain the hits column in the interest_over_time section and add it to a final dataframe that contains a column for each word and the date as index.
I'm a bit lost in how to do this without a for loop
Assuming the gtrends output is the same length for every keyword, you can do the following:
# Load packages
library(purrr)
library(gtrendsR)
# Generate a vector of keywords
words <- c('cruise', 'plane', 'car')
# Download data by iterating gtrends over the vector of keywords
# Extract the hits data and make it into a dataframe for each keyword
trends <- map(.x = words,
~ as.data.frame(gtrends(keyword = .x, time = 'now 1-H')$interest_over_time$hits)) %>%
# Add the keywords as column names to the three dataframes
map2(.x = .,
.y = words,
~ set_names(.x, nm = .y)) %>%
# Convert the list of three dataframes to a single dataframe
map_dfc(~ data.frame(.x))
# Check data
head(trends)
#> cruise plane car
#> 1 50 75 84
#> 2 51 74 83
#> 3 100 67 81
#> 4 46 76 83
#> 5 48 77 84
#> 6 43 75 82
str(trends)
#> 'data.frame': 59 obs. of 3 variables:
#> $ cruise: int 50 51 100 46 48 43 48 53 43 50 ...
#> $ plane : int 75 74 67 76 77 75 73 80 70 79 ...
#> $ car : int 84 83 81 83 84 82 84 87 85 85 ...
Created on 2020-06-27 by the reprex package (v0.3.0)
You can use map to get all the data as a list and use reduce to combine the data.
library(purrr)
library(gtrendsr)
library(dplyr)
map(words, ~gtrends(.x,geo= '', time = 'today 12-m')$interest_over_time %>%
dplyr::select(date, !!.x := hits)) %>%
reduce(full_join, by = 'date')
# date cruise plane car
#1 2019-06-30 64 53 96
#2 2019-07-07 75 48 97
#3 2019-07-14 73 48 100
#4 2019-07-21 74 48 100
#5 2019-07-28 71 47 100
#6 2019-08-04 67 47 97
#7 2019-08-11 68 56 98
#.....
Related
I have a data frame of blood pressure data of the following form:
bpdata <- data.frame(bp1 = c("120/89", "110/70", "121/78"), bp2 = c("130/69", "120/90", "125/72"), bp3 = c("115/90", "112/71", "135/80"))
I would like to use the following extract command, but globally, i.e. on all bp\d columns
extract(bp1, c("systolic_1","diastolic_1"),"(\\d+)/(\\d+)")
How can I capture the digit in the column selection and use it in the column output names? I can hack around this by creating a list of column names and then using one of the apply family, but it seems to me there ought to be a more elegant way to do this.
Any suggestions?
We could use read.csv on multiple columns in a loop (Map) with sep = "/" and cbind the list elements at the end with do.call
do.call(cbind, Map(function(x, y) read.csv(text= x, sep="/", header = FALSE,
col.names = paste0(c('systolic', 'diastolic'), y)),
unname(bpdata), seq_along(bpdata)))
# systolic1 diastolic1 systolic2 diastolic2 systolic3 diastolic3
#1 120 89 130 69 115 90
#2 110 70 120 90 112 71
#3 121 78 125 72 135 80
Or without a loop, paste the columns to a single string for each row and then use read.csv/read.table
read.csv(text = do.call(paste, c(bpdata, sep="/")),
sep="/", header = FALSE,
col.names = paste0(c('systolic', 'diastolic'),
rep(seq_along(bpdata), each = 2)))
# systolic1 diastolic1 systolic2 diastolic2 systolic3 diastolic3
#1 120 89 130 69 115 90
#2 110 70 120 90 112 71
#3 121 78 125 72 135 80
Or using tidyverse, similar option is to unite the column into a single one with /, then use either extract or separate to split the column into multiple columns
library(dplyr)
library(tidyr)
library(stringr)
bpdata %>%
unite(bpcols, everything(), sep="/") %>%
separate(bpcols, into = str_c(c('systolic', 'diastolic'),
rep(seq_along(bpdata), each = 2)), convert = TRUE)
# systolic1 diastolic1 systolic2 diastolic2 systolic3 diastolic3
#1 120 89 130 69 115 90
#2 110 70 120 90 112 71
#3 121 78 125 72 135 80
I have 2880 observations in my data.frame. I have to create a new data.frame in which, I have to select rows from 25-77 from every 96 selected rows.
df.new = df[seq(25, nrow(df), 77), ] # extract from 25 to 77
The above code extracts only row number 25 to 77 but I want every row from 25 to 77 in every 96 rows.
One option is to create a vector of indeces with which subset the dataframe.
idx <- rep(25:77, times = nrow(df)/96) + 96*rep(0:29, each = 77-25+1)
df[idx, ]
You can use recycling technique to extract these rows :
from = 25
to = 77
n = 96
df.new <- df[rep(c(FALSE, TRUE, FALSE), c(from - 1, to - from + 1, n - to))), ]
To explain for this example it will work as :
length(rep(c(FALSE, TRUE, FALSE), c(24, 53, 19))) #returns
#[1] 96
In these 96 values, value 25-77 are TRUE and rest of them are FALSE which we can verify by :
which(rep(c(FALSE, TRUE, FALSE), c(24, 53, 19)))
# [1] 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46
#[23] 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68
#[45] 69 70 71 72 73 74 75 76 77
Now this vector is recycled for all the remaining rows in the dataframe.
First, define a Group variable, with values 1 to 30, each value repeating 96 times. Then define RowWithinGroup and filter as required. Finally, undo the changes introduced to do the filtering.
df <- tibble(X=rnorm(2880)) %>%
add_column(Group=rep(1:96, each=30)) %>%
group_by(Group) %>%
mutate(RowWithinGroup=row_number()) %>%
filter(RowWithinGroup >= 25 & RowWithinGroup <= 77) %>%
select(-Group, -RowWithinGroup) %>%
ungroup()
Welcome to SO. This question may not have been asked in this exact form before, but the proinciples required have been rerefenced in many, many questions and answers,
A one-liner base solution.
lapply(split(df, cut(1:nrow(df), nrow(df)/96, F)), `[`, 25:77, )
Note: Nothing after the last comma
The code above returns a list. To combine all data together, just pass the result above into
do.call(rbind, ...)
I was wondering if there is an easy way to create a table that has the columns as well as row totals?
smoke <- matrix(c(51,43,22,92,28,21,68,22,9),ncol=3,byrow=TRUE)
colnames(smoke) <- c("High","Low","Middle")
rownames(smoke) <- c("current","former","never")
smoke <- as.table(smoke)
I thought this would be super easy, but the solutions i found until now seem to be pretty complicated involving lapply and rbind. However, this seems as such a trivial task, there must be some easier way?
derired results:
> smoke
High Low Middle TOTAL
current 51 43 22 116
former 92 28 21 141
never 68 22 9 99
TOTAL 211 93 52 51
addmargins(smoke)
addmargins is in the stats package.
You can use adorn_totals from janitor :
library(janitor)
library(magrittr)
smoke %>%
as.data.frame.matrix() %>%
tibble::rownames_to_column() %>%
adorn_totals(name = 'TOTAL') %>%
adorn_totals(name = 'TOTAL', where = 'col')
# rowname High Low Middle TOTAL
# current 51 43 22 116
# former 92 28 21 141
# never 68 22 9 99
# TOTAL 211 93 52 356
I have a timeseries with about 100 dates, 50 entities per date (so 5,000 rows) and 50 columns (all are different variables). How can I filter each column in the data frame, per unique date, to keep the top 1/3 of values for each column on each date. Then get the average Return for that group for that date. Thank you.
My data is organized as follows but the numbers in each column are random and vary like they do in column "a" (this is a sample, the real data has many more columns and many more rows):
Date Identity Return a b c d e f... ...z
2/1/19 X 5 75 43 67 85 72 56 92
2/1/19 Y 4 27 43 67 85 72 56 92
2/1/19 Z 7 88 43 67 85 72 56 92
2/1/19 W 2 55 43 67 85 72 56 92
2/2/19 X 7 69 43 67 85 72 56 92
2/2/19 Y 8 23 43 67 85 72 56 92
2/3/19 X 2 34 43 67 85 72 56 92
2/3/19 Y 3 56 43 67 85 72 56 92
2/3/19 Z 4 62 43 67 85 72 56 92
2/3/19 W 4 43 43 67 85 72 56 92
2/3/19 U 4 26 43 67 85 72 56 92
2/4/19 X 6 67 43 67 85 72 56 92
2/4/19 Y 1 78 43 67 85 72 56 92
2/5/19 X 4 75 43 67 85 72 56 92
2/7/19 X 5 99 43 67 85 72 56 92
2/7/19 Y 4 72 43 67 85 72 56 92
2/7/19 Z 4 45 43 67 85 72 56 92
I am trying to filter data into quantiles. I have a code that works for filtering into quantiles for one measure. However I want filtered results for many measures individually (i.e. I want a “high” group for a ton of columns).
The code that I have that works for one measure is as follows.
Columns are date, identity, and a a is the indicator I want to sort on
High = df[!is.na(df$a),] %>%
group_by(df.date) %>%
filter(a > quantile(a, .666)) %>%
summarise(high_return = sum(df.return) / length(df.identity)
Now I want to loop this for when I have many indicators to sort on individually (I.e. I do not want to sort within one another, I want each sorted separately and the results to be broken out by indicator)
I want the output of the loop to be a new data frame with the following format (where a_Return is the average return of the top 1/3 of the original a's on a given date):
Date a_Return b_Return c_Return
2/1/19 6. 7 3
2/3/19 4. 2 5
2/4/19 2. 4 6
I have tried the code below without it working:
Indicators <- c(“a”, “b”, “c”)
for(i in 1:length(Indicators)){
High = df %>%
group_by(df.date) %>%
filter(High[[I]] > quantile(High[[i]], .666)) %>%
summarise(g = sum(df.return) / length(df.identity)}
With this attempt I get the error: "Error in filter_impl(.data, quo) : Result must have length 20, not 4719.
I also tried:
High %>%
group_by(date) %>%
filter_at(vars(Indicators[i]), any_vars(. > quantile (., .666)))%>%
summarise(!!Indicators[I] := sum(Return) / n())
but with that code I get the error "Strings must match column names. Unknown Columns: NA"
I want High to turn up with a date column and then a column for each a, b, and c.
If you combine the filtering and calculations into a single function, then you can put that into summarize_at to apply it easily to each column. Since you're example data isn't fully reproducible, I'll use the iris dataset. In your case, you'd replace Species with Date, and Petal.Width with Return:
library(dplyr)
top_iris <- iris %>%
group_by(Species) %>%
summarize_at(vars(one_of('Sepal.Length', 'Sepal.Width', 'Petal.Length')),
funs(return = sum(Petal.Width[. > quantile(., .666)]) / length(Petal.Width[. > quantile(., .666)])))
top_iris
# A tibble: 3 x 4
Species Sepal.Length_return Sepal.Width_return Petal.Length_return
<fct> <dbl> <dbl> <dbl>
1 setosa 0.257 0.262 0.308
2 versicolor 1.44 1.49 1.49
3 virginica 2.1 2.22 2.09
The problem with using filter is that each function in the pipe runs in order, so any criteria you give to filter_* will have to be applied to the whole data.frame before the result is piped into summarize_at. Instead, we just use a single summarize_at statement, and filter each column as the summarization function is applied to it.
To explain this in more detail, summarize_at takes 2 arguments:
The first argument is one or more of the variable selector functions described in ?select_helpers, enclosed in the vars function. Here we use one_of which just takes a vector of column names, but we could also use matches to select using a regular expession, or starts_with to choose based on a prefix, for example.
The second argument is a list of one or more function calls to be run on each selected column, enclosed in the funs function. Here we have 1 function call, to which we've given the name return.
Like with any tidyverse function, this is evaluated in a local environment constructed from the data piped in. So bare variable names like Petal.Width function as data$Petal.Width. In *_at functions, the . represents the variable passed in, so when the Sepal.Length column is being summarized:
Petal.Width[. > quantile(., .666)]
means:
data$Petal.Width[data$Sepal.Length > quantile(data$Sepal.Length, .666)]
Finally, since the function in funs is named (that's the return =), then the resulting summary columns have the function's name (return) appended to the original column names.
If you want to remove missing data before running these calculations, you can use na.omit to strip out NA values.
To remove all rows containing NA, just pipe your data through na.omit before grouping:
iris2 <- iris
iris2[c(143:149), c(1:2)] <- NA
iris2 %>%
na.omit() %>%
group_by(Species) %>%
summarize_at(vars(one_of('Sepal.Length', 'Sepal.Width', 'Petal.Length')),
funs(return = sum(Petal.Width[. > quantile(., .666)]) / length(Petal.Width[. > quantile(., .666)])))
Species Sepal.Length_return Sepal.Width_return Petal.Length_return
<fct> <dbl> <dbl> <dbl>
1 setosa 0.257 0.262 0.308
2 versicolor 1.44 1.49 1.49
3 virginica 2.09 2.19 2.07
To strip NA values from each column as it's being summarized, you need to move na.omit inside the summarize function:
iris2 %>%
group_by(Species) %>%
summarize_at(vars(one_of('Sepal.Length', 'Sepal.Width', 'Petal.Length')),
funs(return = {
var <- na.omit(.)
length(Petal.Width[var > quantile(var, .666)])
}))
# A tibble: 3 x 4
Species Sepal.Length_return Sepal.Width_return Petal.Length_return
<fct> <dbl> <dbl> <dbl>
1 setosa 0.257 0.262 0.308
2 versicolor 1.44 1.49 1.49
3 virginica 2.11 2.2 2.09
Here we use curly braces to extend the function we run in summarize_at to multiple expressions. First, we strip out NA values, then we calculate the return values. Since this function is in summarize_at it gets applied to each variable based on the grouping established by group_by.
I have a nested loops which produce outputs that I want to store in list objects with dynamic names. A toy example of this would look as follows:
set.seed(8020)
names<-sample(LETTERS,5,replace = F)
for(n in names)
{
#Create the list
assign(paste0("examples_",n),list())
#Poulate the list
get(paste0("examples_",n))[[1]]<-sample(100,10)
get(paste0("examples_",n))[[2]]<-sample(100,10)
get(paste0("examples_",n))[[3]]<-sample(100,10)
}
Unfortunately I keep getting the error:
Error in get(paste0("examples_", n))[[1]] <- sample(100, 10) :
target of assignment expands to non-language object
I have tried all kind of assign, eval, get type of functions to parse the object, but haven't had any luck
Expanding on my comment with a worked example:
examples <- vector(mode="list", length=length(names) )
names(examples) <- names # please change that to mynames
# or almost anything other than `names`
examples <- lapply( examples, function(L) {L[[1]] <- sample(100,10)
L[[2]] <- sample(100,10)
L[[3]] <- sample(100,10); L} )
# Top of the output:
> examples
$P
$P[[1]]
[1] 34 49 6 55 19 28 72 42 14 92
$P[[2]]
[1] 97 71 63 59 66 50 27 45 76 58
$P[[3]]
[1] 94 39 77 44 73 15 51 78 97 53
$F
$F[[1]]
[1] 12 21 89 26 16 93 4 13 62 45
$F[[2]]
[1] 83 21 68 74 32 86 52 49 16 13
$F[[3]]
[1] 14 45 40 46 64 85 88 28 53 42
This mode of programming does become more natural over time. It gets you out of writing clunky for-loops all the time. Develop your algorithms for a single list-node at a time and then use sapply or lapply to iterate the processing.