I have two sources of clinical procedure billing information that I have added together (with rbind). In each row there is a CPT field and a CPT.description field that supplys a brief explanation. However, the descriptions are slightly different from the two sources. I want to be able to combine them. That way, if different words or abbreviations are used, then I can just do a string search to find what I am looking for.
So lets make up a simplified representation of a data table that I was able to generate.
cpt <- c(23456,23456,10000,44555,44555)
description <- c("tonsillectomy","tonsillectomy in >12 year old","brain transplant","castration","orchidectomy")
cpt.desc <- data.frame(cpt,description)
And here is what I want to get to.
cpt.wanted <- c(23456,10000,44555)
description.wanted <- c("tonsillectomy; tonsillectomy in >12 year old","brain transplant","castration; orchidectomy")
cpt.desc.wanted <- data.frame(cpt.wanted,description.wanted)
I have tried using functions such as unstack and then lapply(list,paste) but that is not pasting the elements of each list. I also tried reshape but there was no categorical variable to differentiate first or second version of description or even in some cases a third. The really annoying part is I had a similar problem a few months or years ago and someone helped me either on stackoverflow or on r-help and for the life of me I cannot find it.
So the underlying problem is, imagine that I have a spreadsheet in front of me. I need to do a vertical merge (paste) of two or maybe even three description cells who have the same CPT code in the adjacent column.
What buzzwords should I have been using to search for a solution to this problem.
Thank you so much for your help.
sapply( sapply(unique(cpt), function(x) grep(x, cpt) ),
# creates sets of index vectors as a list
function(x) paste(description[x], collapse=";") )
# ... and this pastes each set of selected items from "description" vector
[1] "tonsillectomy;tonsillectomy in >12 year old"
[2] "brain transplant"
[3] "castration;orchidectomy"
Here is an approach that uses plyr.
library("plyr")
cpt.desc.wanted <- ddply(cpt.desc, .(cpt), summarise,
description.wanted = paste(unique(description), collapse="; "))
which gives
> cpt.desc.wanted
cpt description.wanted
1 10000 brain transplant
2 23456 tonsillectomy; tonsillectomy in >12 year old
3 44555 castration; orchidectomy
Related
I am trying to create another list to include all of the trials from 2 out of the 3 variables shown in the picture.
I am trying to learn how to map this.
So far I have:
d2 <- map(d1 ,`[` , c("time_100L_1", "vertical_100L_1"))
but this only brings in the first trial. I need all 14 for time and vertical and force is in the middle of the list.
any suggestions? See picture for list
map(d1, [, c(paste0("time_100L_", 1:14), paste0("vertical_100L_", 1:14)))
So... I'm very illiterate when it comes to RStudio and I'm using this program for a class... I'm trying to figure out how to sum a subset of a category. I apologize in advance if this doesn't make sense but I'll do my best to explain because I have no clue what I'm doing and would also appreciate an explanation of why and not just what the answer would be. Note: The two lines I included are part of the directions I have to follow, not something I just typed in because I knew how to - I don't... It's the last part, the sum, that I am not explained how to do and thus I don't know what to do and would appreciate help figuring out.
For example,
I have this:
category_name category2_name
1 ABC
2 ABC
3 ABC
4 ABC
5 ABC
6 BDE
5 EFG
7 EFG
I wanted to find the sum of these numbers, so I was told to put in this:
sum(dataname$category_name)
After doing this, I'm asked to type this in, apparently creating a subset.
allabc <- subset(dataname, dataname$category_name2 == "abc")
I created this subset and now I have a new table popped up with this subset. I'm asked to sum only the numbers of this ABC subset... I have absolutely no clue on how to do this. If someone could help me out, I'd really appreciate it!
R is the software you are using. It is case-sensitive. So "abc" is not equal to "ABC".
The arguments are the "things" you put inside functions. Some arguments have the same name as the functions (which is a little confusing at first, but you get used to this eventually). So when I say the subset argument, I am talking about your second argument to the subset function, which you didn't name. That's ok, but when starting to learn R, try to always name your arguments.
So,
allabc <- subset(dataname, dataname$category_name2 == "abc")
Needs to be changed to:
allabc <- subset(dataname, subset=category2_name == "ABC")
And you also don't need to specify the name of the data again in the subset argument, since you've done that already in the first argument (which you didn't name, but almost everyone never bothers to do that).
This is the most easily done using tidyverse.
# Your data
data <- data.frame(category_name = 1:8, category_name2 = c(rep("ABC", 5), "BDE", "EFG", "EFG"))
# Installing tidyverse
install.packages("tidyverse")
# Loading tidyverse
library(tidyverse)
# For each category_name2 the category_name is summed
data %>%
group_by(category_name2) %>%
summarise(sum_by_group = sum(category_name))
# Output
category_name2 sum_by_group
ABC 15
BDE 6
EFG 15
My dataframe column looks like this:
head(tweets_date$Tweet)
[1] b"It is #DineshKarthik's birthday and here's a rare image of the captain of #KKRiders. Have you seen him do this before? Happy birthday, DK\\xf0\\x9f\\x98\\xac
[2] b'The awesome #IPL officials do a wide range of duties to ensure smooth execution of work! Here\\xe2\\x80\\x99s #prabhakaran285 engaging with the #ChennaiIPL kid-squad that wanted to meet their daddies while the presentation was on :) #cutenessoverload #lineofduty \\xf0\\x9f\\x98\\x81
[3] b'\\xf0\\x9f\\x8e\\x89\\xf0\\x9f\\x8e\\x89\\n\\nCHAMPIONS!!
[4] b'CHAMPIONS - 2018 #IPLFinal
[5] b'Chennai are Super Kings. A fairytale comeback as #ChennaiIPL beat #SRH by 8 wickets to seal their third #VIVOIPL Trophy \\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86. This is their moment to cherish, a moment to savour.
[6] b"Final. It's all over! Chennai Super Kings won by 8 wickets
These are tweets which have mentions starting with '#', I need to extract all of them and save each mention in that particular tweet as "#mention1 #mention2". Currently my code just extracts them as lists.
My code:
tweets_date$Mentions<-str_extract_all(tweets_date$Tweet, "#\\w+")
How do I collapse those lists in each row to a form a string separated by spaces as mentioned earlier.
Thanks in advance.
I trust it would be best if you used an asis column in this case:
extract words:
library(stringr)
Mentions <- str_extract_all(lis, "#\\w+")
some data frame:
df <- data.frame(col = 1:6, lett = LETTERS[1:6])
create a list column:
df$Mentions <- I(Mentions)
df
#output
col lett Mentions
1 1 A #DineshK....
2 2 B #IPL, #p....
3 3 C
4 4 D
5 5 E #ChennaiIPL
6 6 F
I think this is better since it allows for quite easy sub setting:
df$Mentions[[1]]
#output
[1] "#DineshKarthik" "#KKRiders"
df$Mentions[[1]][1]
#output
[1] "#DineshKarthik"
and it succinctly shows whats inside the column when printing the df.
data:
lis <- c("b'It is #DineshKarthik's birthday and here's a rare image of the captain of #KKRiders. Have you seen him do this before? Happy birthday, DK\\xf0\\x9f\\x98\\xac",
"b'The awesome #IPL officials do a wide range of duties to ensure smooth execution of work! Here\\xe2\\x80\\x99s #prabhakaran285 engaging with the #ChennaiIPL kid-squad that wanted to meet their daddies while the presentation was on :) #cutenessoverload #lineofduty \\xf0\\x9f\\x98\\x81",
"b'\\xf0\\x9f\\x8e\\x89\\xf0\\x9f\\x8e\\x89\\n\\nCHAMPIONS!!",
"b'CHAMPIONS - 2018 #IPLFinal",
"b'Chennai are Super Kings. A fairytale comeback as #ChennaiIPL beat #SRH by 8 wickets to seal their third #VIVOIPL Trophy \\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86\\xf0\\x9f\\x8f\\x86. This is their moment to cherish, a moment to savour.",
"b'Final. It's all over! Chennai Super Kings won by 8 wickets")
The str_extract_all function from the stringr package returns a list of character vectors. So, if you instead want a list of single CSV terms, then you may try using sapply for a base R option:
tweets <- str_extract_all(tweets_date$Tweet, "#\\w+")
tweets_date$Mentions <- sapply(tweets, function(x) paste(x, collapse=", "))
Demo
Via Twitter's help site: "Your username cannot be longer than 15 characters. Your real name can be longer (20 characters), but usernames are kept shorter for the sake of ease. A username can only contain alphanumeric characters (letters A-Z, numbers 0-9) with the exception of underscores, as noted above. Check to make sure your desired username doesn't contain any symbols, dashes, or spaces."
Note that email addresses can be in tweets as can URLs with #'s in them (and not just the silly URLs with username/password in the host component). Thus, something like:
(^|[^[[:alnum:]_]#/\\!?=&])#([[:alnum:]_]{1,15})\\b
is likely a better, safer choice
I have a list of boroughs and a list of localities (like this one). Each locality lies in exactly one borough. What's the best way to store this kind of hierarchical structure in R, considerung that I'd like to have a convenient and readable way of accessing these, and using this list to accumulate data on the locality-level to the borough level.
I've come up with the following:
localities <- list("Mitte" = c("Mitte", "Moabit", "Hansaviertel", "Tiergarten", "Wedding", "Gesundbrunnen",
"Friedrichshain-Kreuzberg" = c("Friedrichshain", "Kreuzberg")
)
But I am not sure if this is the most elegant and accessible way.
If I wanted to assign additional information on the localitiy-level, I could do that by replacing the c(...) by some other call, like rbind(c('0201', '0202'), c("Friedrichshain", "Kreuzberg")) if I wanted to add additional information to the borough-level (like an abbreviated name and a full name for each list), how would I do this?
Edit: For example, I'd like to condense a table like this into a borough-wise version.
Hard to know without having a better view on how you intend to use this, but I would strongly recommend moving away from a nested list structure to a data frame structure:
library(reshape2)
loc.df <- melt(localities)
This is what the molten data looks like:
value L1
1 Mitte Mitte
2 Moabit Mitte
3 Hansaviertel Mitte
4 Tiergarten Mitte
5 Wedding Mitte
6 Gesundbrunnen Mitte
7 Friedrichshain Friedrichshain-Kreuzberg
8 Kreuzberg Friedrichshain-Kreuzberg
You can then use all the standard data frame and other computations:
loc.df$population <- sample(100:500, nrow(loc.df)) # make up population
tapply(loc.df$population, loc.df$L1, mean) # population by borough
gives mean population by Borough:
Friedrichshain-Kreuzberg Mitte
278.5000 383.8333
For more complex calculations you can use data.table and dplyr
You can extract all of this data directly into a data.frame using the XML library.
library(XML)
theurl <- "http://en.wikipedia.org/wiki/Boroughs_and_localities_of_Berlin#List_of_localities"
tables<-readHTMLTable(theurl)
boroughs<-tables[[1]]$Borough
localities<-tables[c(3:14)]
names(localities) <- as.character(boroughs)
all<-do.call("rbind", localities)
#Roland, I think you will find data frames superior to lists for the reasons cited earlier, but also because there is other data on the web page you reference. Loading to a data frame will make it easy to go further if you wish. For example, making comparisons based on population density or other items provided "for free" on the page will be a snap from a data frame.
I am completely new to R. I tried reading the reference and a couple of good introductions, but I am still quite confused.
I am hoping to do the following:
I have produced a .txt file that looks like the following:
area,energy
1.41155882174e-05,1.0914586287e-11
1.46893363946e-05,5.25011714434e-11
1.39244046855e-05,1.57904991488e-10
1.64155121046e-05,9.0815757601e-12
1.85202830392e-05,8.3207522281e-11
1.5256036289e-05,4.24756620609e-10
1.82107587343e-05,0.0
I have the following command to read the file in R:
tbl <- read.csv("foo.txt",header=TRUE).
producing:
> tbl
area energy
1 1.411559e-05 1.091459e-11
2 1.468934e-05 5.250117e-11
3 1.392440e-05 1.579050e-10
4 1.641551e-05 9.081576e-12
5 1.852028e-05 8.320752e-11
6 1.525604e-05 4.247566e-10
7 1.821076e-05 0.000000e+00
Now I want to store each column in two different vectors, respectively area and energy.
I tried:
area <- c(tbl$first)
energy <- c(tbl$second)
but it does not seem to work.
I need to different vectors (which must include only the numerical data of each column) in order to do so:
> prob(energy, given = area), i.e. the conditional probability P(energy|area).
And then plot it. Can you help me please?
As #Ananda Mahto alluded to, the problem is in the way you are referring to columns.
To 'get' a column of a data frame in R, you have several options:
DataFrameName$ColumnName
DataFrameName[,ColumnNumber]
DataFrameName[["ColumnName"]]
So to get area, you would do:
tbl$area #or
tbl[,1] #or
tbl[["area"]]
With the first option generally being preferred (from what I've seen).
Incidentally, for your 'end goal', you don't need to do any of this:
with(tbl, prob(energy, given = area))
does the trick.