Looking for advice on refining my code and also trimming to a date range.
The spreadsheet itself is pulled from another system and so the structure of the excel cannot be changed. When you pull the data it basically starts at E2, with the first date column in F2, and the first item in E3. The data will continue to populate to the right for as long as it goes on for. I have replicated the structure below.
AndI want it to look like:
I have come up with the below, which works, but I was looking for advice on refining it down to fewer individual step by steps.
In the below code:
= extracting data
= pulling the dates out
= formatting from
excel number to an actual date
= grabbing the item names
= transposing data and skipping some parts
= adding in dates to the row names
#1
df <- data.frame(read_excel("C:/example.xlsx",
sheet = "Sheet1"))
#2
dfdate <- gtb[1, -c(1,2,3,4,5)]
#3
dfdate <- format(as.Date(as.numeric(dfdate),
origin = "1899-12-30"), "%d/%m/%Y")
#4
rownames(gtb) <- gtb[,1]
#5
gtb <- as.data.frame(t(gtb[, -c(1,2,3,4,5)]))
#6
rownames(gtb) <- dfdate
After the row names have been added the structure is such that I am happy to start creating the visuals where needed.
thanks for your advice
David
Here is one suggestion, I don't really have easy access to your data, but I am including code to remove those columns as you do, based on their names, which can be nicer than removing by index.
df <- read.table( text=
"Item_Code 01/01/2018 01/02/2018 01/03/2018 01/04/2018
Item 99 51 60 69
Item2 42 47 88 2
Item3 36 81 42 48
",header=TRUE, check.names=FALSE) %>%
rename( `Item Code` = Item_Code )
library(tibble)
library(lubridate)
x <- df %>% select( -matches("Code \\d|Internal Code") ) %>%
column_to_rownames("Item Code") %>%
t %>% as.data.frame %>%
rownames_to_column("Item Code") %>%
mutate( `Item Code` = dmy(`Item Code`) )
x
Output:
> x
Item Code Item Item2 Item3
1 2018-01-01 99 42 36
2 2018-02-01 51 47 81
3 2018-03-01 60 88 42
4 2018-04-01 69 2 48
I went a bit forth and back with this solution, but it can be nice to also showcase how to remove columns by a regex on their column names, since you are removing several similarly named columns.
The t trick, that you also use, works becuase there is really only one more column there that would cause problems with this, as others have commented, and this can be temporarily stowed away as rownames. If that weren't the case, you're looking at a more complex solution involving pivot_wider and pivot_longer or splitting the data.frame and transposing only one of the halves.
Related
I have a very large df with a column that contains the file directory for each row's data.
Example: D:Mouse_2174/experiment/13/trialsummary.txt.1
I would like to create 2 new columns, one with only the mouse ID (2174) and one with the session number (13). There will be different IDs and session numbers based on the row.
I've used sub as recommended here (match part of names in data.frame to new column), but only can get the subject column to say "D:Mouse_2174" I've added an additional line and can get it down to "D:Mous2174"
Is there a way to eliminate all chars before _ and after / to obtain mouse ID?
For session number, I'm not quite as sure what to do with multiple / in the directory name.
percent_correct_list$mouse_id <- sub("/.+", "", percent_correct_list$rn)
#gives me D:Mouse_2174
percent_correct_list$mouse_id <- sub("+._", "", percent_correct_list$mouse_id)
#gives me D:Mous2174
Here is sample code for the directories:
df <- data.frame(
rn = c("D:Mouse_2174/iti_intervals/9/trialsummary.txt.1",
"D:Mouse_2181/iti_intervals/33/trialsummary.txt.1",
"D:Mouse_2183/iti_intervals/107/trialsummary.txt.2",
"D:Mouse_2185/iti_intervals/87/trialsummary.txt.1")
)
What I want:
rn
id
session
D:..
2174
9
D:..
2181
33
D:..
2183
107
D:..
2185
87
Maybe there's some way to do this earlier along in the process too (like when I import all the data into a df using lapply - but this is good as well)
For sure isnt an elegant solution. Only works if your ID and Session are always numbers...
df <- data.frame(
rn = c("D:Mouse_2174/iti_intervals/9/trialsummary.txt.1",
"D:Mouse_2181/iti_intervals/33/trialsummary.txt.1",
"D:Mouse_2183/iti_intervals/107/trialsummary.txt.2",
"D:Mouse_2185/iti_intervals/87/trialsummary.txt.1")) %>%
# Extract all numeric values from the string
mutate(allnums = regmatches(rn, gregexpr("+[[:digit:]]+", rn)))%>%
# Separate them
separate(allnums, into = c("id", "session", "idk"), sep = "\\,") %>%
# Extract them individually
mutate(id = as.numeric(regmatches(id, gregexpr("+[[:digit:]]+", id,))),
session = as.numeric(regmatches(session, gregexpr("+[[:digit:]]+", session)))) %>%
select(-idk)
Output:
1 D:Mouse_2174/iti_intervals/9/trialsummary.txt.1 2174 9
2 D:Mouse_2181/iti_intervals/33/trialsummary.txt.1 2181 33
3 D:Mouse_2183/iti_intervals/107/trialsummary.txt.2 2183 107
4 D:Mouse_2185/iti_intervals/87/trialsummary.txt.1 2185 87
Here's a somewhat long-winded solution, using tidyr::separate. Perhaps there is something more concise/elegant.
It does assume that all values of rn take the same format.
library(dplyr)
library(tidyr)
new_df <- df %>%
# separate on / into 4 new columns
separate(rn, into = c(paste0("item", 1:4)), sep = "/", remove = FALSE) %>%
# remove unwanted columns
select(-item2, -item4) %>%
# separate again on _ into 2 new columns
separate(item1, sep = "_", into = c("prefix", "id")) %>%
# retain and rename desired columns
select(rn, id, session = item3)
Result:
rn id session
1 D:Mouse_2174/iti_intervals/9/trialsummary.txt.1 2174 9
2 D:Mouse_2181/iti_intervals/33/trialsummary.txt.1 2181 33
3 D:Mouse_2183/iti_intervals/107/trialsummary.txt.2 2183 107
4 D:Mouse_2185/iti_intervals/87/trialsummary.txt.1 2185 87
I am using the pivot function from the lessR package, to create an Excel-like pivot table with two categorical variables that make up the vertical and horizontal categories, and a mean in each cell. (Hope this makes sense).
I followed the code that the documentation (https://cran.r-project.org/web/packages/lessR/vignettes/pivot.html) gives. Let's follow their example:
d <- Read("Employee")
a <- pivot(d, mean, Salary, Dept, Gender)
The data d is like this:
Years Gender Dept Salary JobSat Plan Pre Post
Ritchie, Darnell 7 M ADMN 53788.26 med 1 82 92
Wu, James NA M SALE 94494.58 low 1 62 74
Hoang, Binh 15 M SALE 111074.86 low 3 96 97
Jones, Alissa 5 F <NA> 53772.58 <NA> 1 65 62
Downs, Deborah 7 F FINC 57139.90 high 2 90 86
Afshari, Anbar 6 F ADMN 69441.93 high 2 100 100
Knox, Michael 18 M MKTG 99062.66 med 3 81 84
Campagna, Justin 8 M SALE 72321.36 low 1 76 84
Kimball, Claire 8 F MKTG 61356.69 high 2 93 92
The pivottable a is a nice table, exactly as I want it to look in terms of cell contents, etc. It appears to be a knitr_kable.
Gender F M
Dept
------- --------- ---------
ACCT 63237.16 59626.20
ADMN 81434.00 80963.35
FINC 57139.90 72967.60
MKTG 64496.02 99062.66
SALE 64188.25 86150.97
Next, I would like to make a dataframe out of this, for easier manipulation in my code and for copying it to the clipboard. However, I don't know how to convert a knitr_kable to a dataframe. Here is my code and the error it results in:
as.data.frame(a)
Error in as.data.frame.default(a) :
cannot coerce class ‘"knitr_kable"’ to a data.frame
The knitr-documentation does not say anything about this conversion - it is only about converting a dataframe to a knitr_kable, which is the opposite of what I want.
I have also tried pivottabler, but this has similar issues: the resulting class cannot be coerced to a dataframe either.
Here are two potential answers:
Most direct: Wrangle the data yourself
If you're open to a tidyverse-style approach, it only takes a few lines to do the wrangling and summarising yourself. That will give you a datatable output that you can work with right away.
# load packages
library(lessR)
library(dplyr)
library(tidyr)
# load data
d <- Read("Employee")
# use tidyverse-style code to pivot and summarise the data yourself
d %>%
group_by(Gender, Dept) %>%
summarise(Salary_mean = mean(Salary)) %>%
pivot_wider(names_from= "Gender", values_from = "Salary_mean")
Read the knitr::kable() markdown output into a data frame
If you prefer to work backwards from a knitr::kable() output to a dataframe, this is addressed in this SO question: Markdown table to data frame in R
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 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.
Can anyone help arranging long data to wide data, but complicated by linked results, ie listed in wide format identified by Study Number this repetitive results listed in wide format after the SN (I've shown an abbreviated table there are more results per patient listed along the bottom with repetitive in columns LabTest, LabDate, Result, Lower, Upper)...I've tried melt and recast, and binding columns but can't seem to get it to work. Over 1000 results to reformat so can't input results manually need to reformat a long data excel document in wide format in R Thank you
Original data looks like this
SN LabTest LabDate Result Lower Upper
TD62 Creat 05/12/2004 22 30 90
TD62 AST 06/12/2004 652 6 45
TD58 Creat 26/05/2007 72 30 90
TD58 Albumin 26/05/2005 22 25 35
TD14 AST 28/02/2007 234 6 45
TD14 Albumin 26/02/2007 15 25 35
Formatted data should look like this
SN LabTCode LabDate Result Lower Upper LabCode LabDate Result Lower Upper
TD62 Creat 05/12/04 22 30 90 AST 06/12/04 652 6 45
TD58 Creat 26/05/05 72 30 90 Alb 26/05/05 22 25 35
TD14 AST 28/02/07 92 30 90 Alb 26/02/07 15 25 35
Formatted data looks like this
So far I have tried:
data_wide2 <- dcast(tdl, SN + LabDate ~ LabCode, value.var="Result")
and
melt(tdl, id = c("SN", "LabDate"), measured= c("Result", "Upper", + "Lower"))
Your issue is that R won't like the final table because it has duplicate column names. Maybe you need the data in that format but it's a bad way to store data because it would be difficult to put the columns back into rows again without a load of manual work.
That said, if you want to do it you'll need a new column to help you transpose the data.
I've used dplyr and tidyr below, which are worth looking at rather than reshape. They're by the same author but more modern and designed to fit together as part of the 'tidyverse'.
library(dplyr)
library(tidyr)
#Recreate your data (not doing this bit in your question is what got you downvoted)
df <- data.frame(
SN = c("TD62","TD62","TD58","TD58","TD14","TD14"),
LabTest = c("Creat","AST","Creat","Albumin","AST","Albumin"),
LabDate = c("05/12/2004","06/12/2004","26/05/2007","26/05/2005","28/02/2007","26/02/2007"),
Result = c(22,652,72,22,234,15),
Lower = c(30,6,30,25,6,25),
Upper = c(90,45,90,35,45,35),
stringsAsFactors = FALSE
)
output <- df %>%
group_by(SN) %>%
mutate(id_number = row_number()) %>% #create an id number to help with tracking the data as it's transposed
gather("key", "value", -SN, -id_number) %>% #flatten the data so that we can rename all the column headers
mutate(key = paste0("t",id_number, key)) %>% #add id_number to the column names. 't' for 'test' to start name with a letter.
select(-id_number) %>% #don't need id_number anymore
spread(key, value)
SN t1LabDate t1LabTest t1Lower t1Result t1Upper t2LabDate t2LabTest t2Lower t2Result t2Upper
<chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
1 TD14 28/02/2007 AST 6 234 45 26/02/2007 Albumin 25 15 35
2 TD58 26/05/2007 Creat 30 72 90 26/05/2005 Albumin 25 22 35
3 TD62 05/12/2004 Creat 30 22 90 06/12/2004 AST 6 652 45
And you're there, possibly with some sorting issues still to crack if you need the columns in a specific order.