I'm doing some very basic sentiment analysis on a pretty large set of data that continues to grow every day. I need to feed this data into a shiny app where I can adjust the date range. Rather than running the analysis over and over again, what I'd like to do is create a new CSV with the sum of each sentiment score by date. I'm having trouble iterating over the date though. Here's some sample data and the lapply() statement I tried that is not working.
library(tidyverse)
library(syuzhet)
library(data.table)
df <- data.frame(date = c("2021-01-18", "2021-01-18", "2021-01-18", "2021-01-17","2021-01-17", "2021-01-16", "2021-01-15", "2021-01-15", "2021-01-15"),
text = c("Some text here", "More text", "Some other words", "Just making this up", "as I go along", "hope the example helps", "thank you in advance", "I appreciate the help", "the end"))
> df
date text
1 2021-01-18 Some text here
2 2021-01-18 More text
3 2021-01-18 Some other words
4 2021-01-17 Just making this up
5 2021-01-17 as I go along
6 2021-01-16 hope the example helps
7 2021-01-15 thank you in advance
8 2021-01-15 I appreciate the help
9 2021-01-15 the end
dates_scores_df <- lapply(df, function(i){
data <- df %>%
# Filter to the unique date
filter(date == unique(df$date[i]))
# Sentiment Analysis for each date
sentiment_data <- get_nrc_sentiment(df$text)
# Convert to df
score_df <- data.frame(sentiment_data[,])
# Transpose the data frame and adjust column names
daily_sentiment_data <- transpose(score_df)
colnames(daily_sentiment_data) <- rownames(score_df)
# Add a date column
daily_sentiment_data$date <- df$date[i]
})
sentiment_scores_by_date <- do.call("rbind.data.frame", dates_scores_df)
What I'd like to get to is something like this (data here is made up and will not match the example above)
date anger anticipation disgust fear joy sadness surprise trust negative positive
2021-01-18 1 2 0 1 2 0 2 1 1 2
2021-01-17 1 2 0 2 3 3 1 2 0 1
You can try :
library(dplyr)
library(purrr)
library(syuzhet)
df %>%
split(.$date) %>%
imap_dfr(~get_nrc_sentiment(.x$text) %>%
summarise(across(.fns = sum)) %>%
mutate(date = .y, .before = 1)) -> result
result
Function lapply iterates over elements of a list. Data frame is technically a list with each column as an element of that list. So in your example you are iterating over columns rather than rows, or even dates (that seems to be your goal). Instead of lapply I'd use dplyr::group_by in combination with one of: dplyr::do, dplyr::summarize or tidyr::nest. See documentations for each function to figure out which function suits the most your need.
Related
I created this function to count the maximum number of consecutive characters in a word.
max(rle(unlist(strsplit("happy", split = "")))$lengths)
The function works on individual words, but when I try to use the function within a mutate step it doesn't work. Here is the code that involves the mutate step.
text3 <- "The most pressing of those issues, considering the franchise's
stated goal of competing for championships above all else, is an apparent
disconnect between Lakers vice president of basketball operations and general manager"
text3_df <- tibble(line = 1:1, text3)
text3_df %>%
unnest_tokens(word, text3) %>%
mutate(
num_letters = nchar(word),
num_vowels = get_count(word),
num_consec_char = max(rle(unlist(strsplit(word, split = "")))$lengths)
)
The variables num_letters and num_vowels work fine, but I get a 2 for every value of num_consec_char. I can't figure out what I'm doing wrong.
This command rle(unlist(strsplit(word, split = "")))$lengths is not vectorized and thus is operating on the entire list of words for each row thus the same result for each row.
You will need to use some type of loop (ie for, apply, purrr::map) to solve it.
library(dplyr)
library(tidytext)
text3 <- "The most pressing of those issues, considering the franchise's
stated goal of competing for championships above all else, is an apparent
disconnect between Lakers vice president of basketball operations and general manager"
text3_df <- tibble(line = 1:1, text3)
output<- text3_df %>%
unnest_tokens(word, text3) %>%
mutate(
num_letters = nchar(word),
# num_vowels = get_count(word),
)
output$num_consec_char<- sapply(output$word, function(word){
max(rle(unlist(strsplit(word, split = "")))$lengths)
})
output
# A tibble: 32 × 4
line word num_letters num_consec_char
<int> <chr> <int> <int>
1 1 the 3 1
2 1 most 4 1
3 1 pressing 8 2
4 1 of 2 1
5 1 those 5 1
6 1 issues 6 2
7 1 considering 11 1
I'm working with R for the first time for a class in college. To preface this: I don't know enough to know what I don't know, so I'm sorry if this question has been asked before. I am trying to predict the results of the Texas state house elections in 2020, and I think the best prior for that is the results of the 2018 state house elections. There are 150 races, so I can't bare to input them all by hand, but I can't find any spreadsheet that has data formatted how I want it. I want it in a pretty standard table format:
My desired table format. However, the table from the Secretary of state I have looks like the following:
Gross ugly table.
I wrote some psuedo code:
Here's the Psuedo Code, basically we want to construct a new CSV:
'''%First, we want to find a district, the house races are always preceded by a line of dashes, so I will need a function like this:
Create a New CSV;
for(x=1; x<151 ; x +=1){
Assign x to the cell under the district number cloumn;
Find "---------------" ;
Go down one line;
Go over two lines;
% We should now be in the third column and now want to read in which party got how many votes. The number of parties is not consistant, so we need to account for uncontested races, libertarians, greens, and write ins. I want totals for Republicans, Democrats, and Other.
while(cell is not empty){
Party <- function which reads cell (but I want to read a string);
go right one column;
Votes <- function which reads cell (but I want to read an integer);
if(Party = Rep){
put this data in place in new CSV;
else if (Party = Dem)
put this data in place in new CSV;
else
OtherVote += Votes;
};
};
Assign OtherVote to the column for other party;
OtherVote <- 0;
%Now I want to assign 0 to null cells (ones where no rep, or no Dem, or no other party contested
read through single row 4 spaces, if its null assign it 0;
Party <- null
};'''
But I don't know enough to google what to do! Here's what I need help with: Can I create a new CSV in Rstudio, how? How can I read specific cells in a table, hopefully indexing? Lastly, how do I write to a table in R. Any help is appreciated! Thank you!
Can I create a new CSV in Rstudio, how?
Yes you can. Use the "write.csv" function.
write.csv(df, file = "df.csv") #see help for more information.
How can I read specific cells in a table?
Use the brackets after df,example below.
df <- data.frame(x = c(1,2,3), y = c("A","B","C"), z = c(15,25,35))
df[1,1]
#[1] 1
df[1,1:2]
# x y
#1 1 A
How do I write to a table in R?
If you want to write a table in xlsx use the function write.xlsx from openxlsx package.
Wikipedia seems to have a table that is closer to the format you are looking for.
In order to get to the table you are looking for we need a few steps:
Download data from Wikipedia and extract table.
Clean up table.
Select columns.
Calculate margins.
1. Download data from wikipedia and extract table.
The rvest table helps with downloading and parsing websites into R objects.
First we download the HTML of the whole website.
library(dplyr)
library(rvest)
wiki_html <-
read_html(
"https://en.wikipedia.org/wiki/2018_United_States_House_of_Representatives_elections_in_Texas"
)
There are a few ways to get a specific object from an HTML file in this case
I dedided to look for the table that has the class name “wikitable plainrowheaders sortable”,
as I learned from inspecting the code, that the only table with that class is
the one we want to extract.
library(purrr)
html_nodes(wiki_html, "table") %>%
map_lgl( ~ html_attr(., "class") == "wikitable plainrowheaders sortable") %>%
which()
#> [1] 20
Then we can select table number 20 and convert it to a dataframe with html_table()
raw_table <-
html_nodes(wiki_html, "table")[[20]] %>%
html_table(fill = TRUE)
2. Clean up table.
The table has duplicated names, we can change that by using as_tibble() and its .name_repair argument. We then usedplyr::select() to get the columns. Furthermore we usedplyr::filter() to delete the first two rows, that have "District" as a value in theDistrictcolumn. Now the columns are still characters
vectors, but we need them to be numeric, therefore we first delete commas from
all columns and then transform columns 2 to 4 to numeric.
clean_table <-
raw_table %>%
as_tibble(.name_repair = "unique") %>%
filter(District != "District") %>%
mutate_all( ~ gsub(",", "", .)) %>%
mutate_at(2:4, as.numeric)
3. Select columns and 4. Calculate margins.
We use dplyr::select() to select the columns you are interested in and give them more helpful names.
Finally we calculate the margin between democratic and republican votes by first adding up there votes
as total_votes and then dividing the difference by total_votes.
clean_table %>%
select(District,
RepVote = Republican...2,
DemVote = Democratic...4,
OthVote = Others...6) %>%
mutate(
total_votes = RepVote + DemVote,
margin = abs(RepVote - DemVote) / total_votes * 100
)
#> # A tibble: 37 x 6
#> District RepVote DemVote OthVote total_votes margin
#> <chr> <dbl> <dbl> <chr> <dbl> <dbl>
#> 1 District 1 168165 61263 3292 229428 46.6
#> 2 District 2 139188 119992 4212 259180 7.41
#> 3 District 3 169520 138234 4604 307754 10.2
#> 4 District 4 188667 57400 3178 246067 53.3
#> 5 District 5 130617 78666 224 209283 24.8
#> 6 District 6 135961 116350 3731 252311 7.77
#> 7 District 7 115642 127959 0 243601 5.06
#> 8 District 8 200619 67930 4621 268549 49.4
#> 9 District 9 0 136256 16745 136256 100
#> 10 District 10 157166 144034 6627 301200 4.36
#> # … with 27 more rows
Edit: In case you want to go with the data provided by the state, it looks to me as if the data you are looking for is in the first, third and fourth column. So what you want to do is.
(All the code below is not tested, as I do not have the original data.)
read data into R
library(readr)
tx18 <- read_csv("filename.csv")
select relevant columns
tx18 <- tx18 %>%
select(c(1,3,4))
clean table
tx18 <- tx18 %>%
filter(!is.na(X3),
X3 != "Party",
X3 != "Race Total")
Group and summarize data by party
tx18 <- tx18 %>%
group_by(X3) %>%
summarise(votes = sum(X3))
Pivot/ Reshape data to wide format
tx18 %>$
pivot_wider(names_from = X3,
values_from = votes)
After this you could then calculate the margin similarly as I did with the Wikipedia data.
I have a dataframe of tweets in R, looking like this:
tweet_text tweet_time rdate twt
<chr> <dttm> <date> <dbl>
1 No New England cottage is complete without nautical t.. 2016-08-25 09:21:00 2016-08-25 1
2 Justice Scalia spent his last hours with members of co… 2016-11-24 16:28:00 2016-11-24 1
3 WHAT THE FAILED OKLAHOMA ABORTION BILL TELLS US http:/… 2016-11-24 16:27:00 2016-11-24 1
4 Bipartisan bill in US Senate to restrict US arms sales… 2016-10-26 07:03:00 2016-10-26 1
5 #MustResign campaign is underway with the heat p his S… 2016-10-01 08:15:00 2016-10-01 1
Each tweet has a specific date assigned, all tweets in the dataframe are from a period of one year. I want to find out a frequency of one specific word ("Senate" for example) over the entire period and plot a graph capturing how the frequency changed over time. I am fairly new to R and I could only think of super complicated ways to do it, but I am sure there must be some that's really easy and simple.
I appreciate any suggestions.
textFreq <- function(pattern, text){
freq <- gregexpr(pattern = pattern, text = text, ignore.case = TRUE)
freq <- lapply(freq, FUN = function(x){
if(length(x)==1&&x==-1){
return(0)
} else {
return(length(x))
}
})
freq <- unlist(freq)
return(freq)
}
test.text <- c("senate.... SENate.. sen","Working in the senate...", "I like dogs")
textFreq(pattern = "senate", test.text)
# [1] 2 1 0
you can use dplyr to group by time periods and use mutate
library(dplyr)
library(magrittr)
data <- data %>%
group_by(*somedatefactor*) %>% #if you wanted to aggrigate every 10 days or something
mutate(SenateFreqPerTweet = textFreq(pattern = "Senate", text = tweet_text),
SenateFreqTotal = sum(SenateFreqPerTweet)) #Counts sum based on current grouping
You may even wrap the previous statement into another function. To do so check out programming with dplyr
But regardless, using this approach you can easily plot the SenateFreqTotal with ggplot2 package
data2 <- data %>% #may be helpful to reduce the size of the dataframe before plotting.
select(SenateFreqTotal, *somedatefactor*) %>%
distinct()
ggplot(data2, aes(y=SenateFreqTotal, x = *somedatefactor*)+ geom_bar(stat="identity")
if you do not want to aggregate the frequencies you can just plot like so
ggplot(data, aes(y=SenateFreqPerTweet, x = tweet_time)) +
geom_bar(stat = "identity")
I got an XLSX with data from a questionnaire for my master thesis.
The questions and answers for an interviewee are in one row in the second column. The first column contains the date.
The data of the second column comes in a form like this:
"age":"52","height":"170","Gender":"Female",...and so on
I started with:
test12 <- read_xlsx("Testdaten.xlsx")
library(splitstackshape)
test13 <- concat.split(data = test12, split.col= "age", sep =",")
Then I got the questions and the answers as a column divided by a ":".
For e.g. column 1: "age":"52" and column2:"height":"170".
But the data is so messy that sometimes in the column of the age question and answer there is a height question and answer and for some questionnaires questions and answers double.
I would need the questions as variables and the answers as observations. But I have no clue how to get there. I could clean the data in excel first, but with the fact that columns are not constant and there are for e.g. some height questions in the age column I see no chance to do it as I will get new data regularly, formated the same way.
Here is an example of the data:
A tibble: 5 x 2
partner.createdAt partner.wphg.info
<chr> <chr>
1 2019-11-09T12:13:11.099Z "{\"age_years\":\"50\",\"job_des\":\"unemployed\",\"height_cm\":\"170\",\"Gender\":\"female\",\"born_in\":\"Italy\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"200000\""
2 2019-11-01T06:43:22.581Z "{\"age_years\":\"34\",\"job_des\":\"self-employed\",\"height_cm\":\"158\",\"Gender\":\"male\",\"born_in\":\"Germany\",\"Alcoholic\":\"true\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"10000\""
3 2019-11-10T07:59:46.136Z "{\"age_years\":\"24\",\"height_cm\":\"187\",\"Gender\":\"male\",\"born_in\":\"England\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"150000\""
4 2019-11-11T13:01:48.488Z "{\"age_years\":\"59\",\"job_des\":\"employed\",\"height_cm\":\"167\",\"Gender\":\"female\",\"born_in\":\"United States\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"2\",\"total_wealth\":\"1000000~
5 2019-11-08T14:54:26.654Z "{\"age_years\":\"36\",\"height_cm\":\"180\",\"born_in\":\"Germany\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"170000\",\"job_des\":\"employed\",\"Gender\":\"male\""
Thank you so much for your time!
You can loop through each entry, splitting at , as you did. Then you can loop through them all again, splitting at :.
The result will be a bunch of variable/value pairings. This can be all done stacked. Then you just want to pivot back into columns.
data
Updated the data based on your edit.
data <- tribble(~partner.createdAt, ~partner.wphg.info,
'2019-11-09T12:13:11.099Z', '{\"age_years\":\"50\",\"job_des\":\"unemployed\",\"height_cm\":\"170\",\"Gender\":\"female\",\"born_in\":\"Italy\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"200000\"',
'2019-11-01T06:43:22.581Z', '{\"age_years\":\"34\",\"job_des\":\"self-employed\",\"height_cm\":\"158\",\"Gender\":\"male\",\"born_in\":\"Germany\",\"Alcoholic\":\"true\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"10000\"',
'2019-11-10T07:59:46.136Z', '{\"age_years\":\"24\",\"height_cm\":\"187\",\"Gender\":\"male\",\"born_in\":\"England\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"3\",\"total_wealth\":\"150000\"',
'2019-11-11T13:01:48.488Z', '{\"age_years\":\"59\",\"job_des\":\"employed\",\"height_cm\":\"167\",\"Gender\":\"female\",\"born_in\":\"United States\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"2\",\"total_wealth\":\"1000000\"',
'2019-11-08T14:54:26.654Z', '{\"age_years\":\"36\",\"height_cm\":\"180\",\"born_in\":\"Germany\",\"Alcoholic\":\"false\",\"knowledge_selfass\":\"5\",\"total_wealth\":\"170000\",\"job_des\":\"employed\",\"Gender\":\"male\"')
libraries
We need a few here. Or you can just call tidyverse.
library(stringr)
library(purrr)
library(dplyr)
library(tibble)
library(tidyr)
function
This function will create a data frame (or tibble) for each question. The first column is the date, the second is the variable, the third is the value.
clean_record <- function(date, text) {
clean_records <- str_split(text, pattern = ",", simplify = TRUE) %>%
str_remove_all(pattern = "\\\"") %>% # remove double quote
str_remove_all(pattern = "\\{|\\}") %>% # remove curly brackets
str_split(pattern = ":", simplify = TRUE)
tibble(date = as.Date(date), variable = clean_records[,1], value = clean_records[,2])
}
iteration
Now we use pmap_dfr from purrr to loop over the rows, outputting each row with an id variable named record.
This will stack the data as described in the function. The mutate() line converts all variable names to lowercase. The distinct() line will filter out rows that are exact duplicates.
What we do then is just pivot on the variable column. Of course, replace data with whatever you name your data frame.
data_clean <- pmap_dfr(data, ~ clean_record(..1, ..2), .id = "record") %>%
mutate(variable = tolower(variable)) %>%
distinct() %>%
pivot_wider(names_from = variable, values_from = value)
result
The result is something like this. Note how I had reordered some of the columns, but it still works. You are probably not done just yet. All columns are now of type character. You need to figure out the desired type for each and convert.
# A tibble: 5 x 10
record date age_years job_des height_cm gender born_in alcoholic knowledge_selfass total_wealth
<chr> <date> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 1 2019-11-09 50 unemployed 170 female Italy false 5 200000
2 2 2019-11-01 34 self-employed 158 male Germany true 3 10000
3 3 2019-11-10 24 NA 187 male England false 3 150000
4 4 2019-11-11 59 employed 167 female United States false 2 1000000
5 5 2019-11-08 36 employed 180 male Germany false 5 170000
For example, convert age_years to numeric.
data_clean %>%
mutate(age_years = as.numeric(age_years))
I am sure you may run into other things, but this should be a start.
I'm new to R, so please go easy on me... I have some longitudinal data that looks like
Basically, I'm trying to find a way to get a table with a) the number of unique cases that have all complete data and b) the number of unique cases that have at least one incomplete or missing data. The end results would ideally be
df<- df %>% group_by(Location)
df1<- df %>% group_by(any(Completion_status=='Incomplete' | 'Missing'))
Not sure about what you want, because it seems there are something of inconsistent between your request and the desired output, however lets try, it seems you need a kind of frequency table, that you can manage with basic R. At the bottom of the answer you can find some data similar to yours.
# You have two cases, the Complete, and the other, so here a new column about it:
data$case <- ifelse(data$Completion_status =='Complete','Complete', 'MorIn')
# now a frequency table about them: if you want a data.frame, here we go
result <- as.data.frame.matrix(table(data$Location,data$case))
# now the location as a new column rather than the rownames
result$Location <- rownames(result)
# and lastly a data.frame with the final results: note that you can change the names
# of the columns but if you want spaces maybe a tibble is better
result <- data.frame(Location = result$Location,
`Number.complete` = result$Complete,
`Number.incomplete.missing` = result$MorIn)
result
Location Number.complete Number.incomplete.missing
1 London 0 1
2 Los Angeles 0 1
3 Paris 3 1
4 Phoenix 0 2
5 Toronto 1 1
Or if you prefere a dplyr chain:
data %>%
mutate(case = ifelse(data$Completion_status =='Complete','Complete', 'MorIn')) %>%
do( as.data.frame.matrix(table(.$Location,.$case))) %>%
mutate(Location = rownames(.)) %>%
select(3,1,2) %>%
`colnames<-`(c("Location","Number of complete ", "Number of incomplete or"))
Location Number of complete Number of incomplete or
1 London 0 1
2 Los Angeles 0 1
3 Paris 3 1
4 Phoenix 0 2
5 Toronto 1 1
With data:
# here your data (next time try to put them in an usable way in the question)
data <- data.frame( ID = c("A1","A1","A2","A2","B1","C1","C2","D1","D2","E1"),
Location = c('Paris','Paris','Paris','Paris','London','Toronto','Toronto','Phoenix','Phoenix','Los Angeles'),
Completion_status = c('Complete','Complete','Incomplete','Complete','Incomplete','Missing',
'Complete','Incomplete','Incomplete','Missing'))