Combine separate column/rows as one column/row in R - r

using txt.file, i have this dataset:
Xenopsylla cheopis Echinolaelaps sp.
Maxomys rajah 1 3
Callosciurus prevostii borneensis 4 2
using this function,
test<-read.table("data.txt",header=T)
Xenopsylla cheopis Echinolaelaps sp.
Maxomys rajah 1 3
Callosciurus prevostii borneensis 4 2
R seems to recognize my data as different columns/rows and produce this error:
Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec, :
line 2 did not have 4 elements
i tried to use textConnection but it seems that it does not produce what i want


First of all just store your data in a character vector as I did here:
test<-readChar("C:/Users/Julian/Downloads/file.txt", file.info("C:/Users/Julian/Downloads/file.txt")$size)
Obviously, you need to replace the path of my file with yours.
Then you get rid of the space between Genus and Species using gsub()
test<-gsub("([[:lower:]])([[:space:]])([[:lower:]])", "\\1\\3",test)
Finally, you can read your data using read.table() with the text argument:
a<-read.table(text=test,sep="\t",header=TRUE,row.names = 1)
a
Xenopsyllacheopis Echinolaelapssp. Ixodessp.
Maxomysrajah 3 8 9
Callosciurusprevostiiborneensis 5 7 1
Sundamysmuelleri 3 5 7
Niviventercremoriventer 6 8 9
EDIT:
To answer OP's new question in the comments:
"([[:lower:]])([[:space:]])([[:lower:]])"
enables us to find all the parts of the strings that we created with readChar() that match this pattern. This pattern is: a lowercase letter followed by a blank space followed by a lowercase letter.
You can understand this match the genus and species name but not a species name and the following genus because a genus starts with an uppercase letter.
Now the "\\1\\3" part means that we keep the first and third part of our
"([[:lower:]])([[:space:]])([[:lower:]])" pattern. That is ([[:lower:]]) and ([[:lower:]]). Because there is no space between "\\1 and \\3 in "\\1\\3" we will join them without spaces. Therefore we will have Genusspecies instead of Genus species.

Related

Grepl for 2 words/phrases in proximity in R (dplyr)

I'm trying to create a filter for large dataframe. I'm trying to use grepl to search for a series of text within a specific column. I've done this for single words/combinations, but now I want to search for two words in close proximity (ie the word tumo(u)r within 3 words of the word colon).
I've checked my regular expression on https://www.regextester.com/109207 and my search works there, but it doesn't work within R.
The error I get is
Error: '\W' is an unrecognized escape in character string starting ""\btumor|tumour)\W"
Example below - trying to search for tumo(u)r within 3 words of cancer.
Can anyone help?
library(tibble)
example.df <- tibble(number = 1:4, AB = c('tumor of the colon is a very hard disease to cure', 'breast cancer is also known as a neoplasia of the breast', 'tumour of the colon is bad', 'colon cancer is also bad'))
filtered.df <- example.df %>%
filter(grepl(("\btumor|tumour)\W|\w+(\w+\W+){0,3}colon\b"), AB, ignore.case=T)

R uses backslashes as escapes and the regex engine does,too. Need to double your backslashes. This is explained in multiple prior questions on StackOverflow as well as in the help page brought up at ?regex. You should try to use the escaped operators in a more simple set of tests before attempting complex operations. And you should pay better attention to the proper placement of parentheses and quotes in the pattern argument.
filtered.df <- example.df %>%
#filter(grepl(("\btumor|tumour)\W|\w+(\w+\W+){0,3}colon\b"), AB,
# errors here ....^.^..............^..^...^..^.............^.^
filter(grepl( "(\\btumor|tumour)\\W|\\w+(\\w+\\W+){0,3}colon\\b", AB,
ignore.case=T) )
> filtered.df
# A tibble: 2 × 2
number AB
<int> <chr>
1 1 tumor of the colon is a very hard disease to cure
2 3 tumour of the colon is bad

regex: extract segments of a string containing a word, between symbols

Hello I have a data frame that looks something like this
dataframe <- data_frame(text = c('WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12',
'WUFF;other stuff to keep;WIFF2;yes yes IGWIFF'))
print(dataframe)
# A tibble: 2 × 1
text
<chr>
1 WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12
2 WUFF;other stuff to keep;WIFF2;yes yes IGWIFF
I want to extract the segment of the strings containing the word "keep". Note that these segments can be separated from other parts by different symbols for example , and ;.
the final dataset should look something like this.
final_dataframe <- data_frame(text = c('some words to keep',
'other stuff to keep'))
print(final_dataframe)
# A tibble: 2 × 1
text
<chr>
1 some words to keep
2 other stuff to keep
Does anyone know how I could do this?

With stringr ...
library(stringr)
library(dplyr)
dataframe %>%
mutate(text = trimws(str_extract(text, "(?<=[,;]).*keep")))
# A tibble: 2 × 1
text
<chr>
1 some words to keep
2 other stuff to keep
Created on 2022-02-01 by the reprex package (v2.0.1)

I've made great use of the positive lookbehind and positive lookahead group constructs -- check this out: https://regex101.com/r/Sc7h8O/1
If you want to assert that the text you're looking for comes after a character/group -- in your first case the apostrophe, use (?<=').
If you want to do the same but match something before ' then use (?=')
And you want to match between 0 and unlimited characters surrounding "keep" so use .* on either side, and you wind up with (?<=').*keep.*(?=')
I did find in my test that a string like text =' c('WAFF, some words to keep, ciao, WOFF hey ;other ;WIFF12', will also match the c(, which I didn't intend. But I assume your strings are all captured by pairs of apostrophes.

Find the most frequently occuring words in a text in R

Can someone help me with how to find the most frequently used two and three words in a text using R?
My text is...
text <- c("There is a difference between the common use of the term phrase and its technical use in linguistics. In common usage, a phrase is usually a group of words with some special idiomatic meaning or other significance, such as \"all rights reserved\", \"economical with the truth\", \"kick the bucket\", and the like. It may be a euphemism, a saying or proverb, a fixed expression, a figure of speech, etc. In grammatical analysis, particularly in theories of syntax, a phrase is any group of words, or sometimes a single word, which plays a particular role within the grammatical structure of a sentence. It does not have to have any special meaning or significance, or even exist anywhere outside of the sentence being analyzed, but it must function there as a complete grammatical unit. For example, in the sentence Yesterday I saw an orange bird with a white neck, the words an orange bird with a white neck form what is called a noun phrase, or a determiner phrase in some theories, which functions as the object of the sentence. Theorists of syntax differ in exactly what they regard as a phrase; however, it is usually required to be a constituent of a sentence, in that it must include all the dependents of the units that it contains. This means that some expressions that may be called phrases in everyday language are not phrases in the technical sense. For example, in the sentence I can't put up with Alex, the words put up with (meaning \'tolerate\') may be referred to in common language as a phrase (English expressions like this are frequently called phrasal verbs\ but technically they do not form a complete phrase, since they do not include Alex, which is the complement of the preposition with.")

The tidytext package makes this sort of thing pretty simple:
library(tidytext)
library(dplyr)
data_frame(text = text) %>%
unnest_tokens(word, text) %>% # split words
anti_join(stop_words) %>% # take out "a", "an", "the", etc.
count(word, sort = TRUE) # count occurrences
# Source: local data frame [73 x 2]
#
# word n
# (chr) (int)
# 1 phrase 8
# 2 sentence 6
# 3 words 4
# 4 called 3
# 5 common 3
# 6 grammatical 3
# 7 meaning 3
# 8 alex 2
# 9 bird 2
# 10 complete 2
# .. ... ...
If the question is asking for counts of bigrams and trigrams, tokenizers::tokenize_ngrams is useful:
library(tokenizers)
tokenize_ngrams(text, n = 3L, n_min = 2L, simplify = TRUE) %>% # tokenize bigrams and trigrams
as_data_frame() %>% # structure
count(value, sort = TRUE) # count
# Source: local data frame [531 x 2]
#
# value n
# (fctr) (int)
# 1 of the 5
# 2 a phrase 4
# 3 the sentence 4
# 4 as a 3
# 5 in the 3
# 6 may be 3
# 7 a complete 2
# 8 a phrase is 2
# 9 a sentence 2
# 10 a white 2
# .. ... ...

Your text is:
text <- c("There is a difference between the common use of the term phrase and its technical use in linguistics. In common usage, a phrase is usually a group of words with some special idiomatic meaning or other significance, such as \"all rights reserved\", \"economical with the truth\", \"kick the bucket\", and the like. It may be a euphemism, a saying or proverb, a fixed expression, a figure of speech, etc. In grammatical analysis, particularly in theories of syntax, a phrase is any group of words, or sometimes a single word, which plays a particular role within the grammatical structure of a sentence. It does not have to have any special meaning or significance, or even exist anywhere outside of the sentence being analyzed, but it must function there as a complete grammatical unit. For example, in the sentence Yesterday I saw an orange bird with a white neck, the words an orange bird with a white neck form what is called a noun phrase, or a determiner phrase in some theories, which functions as the object of the sentence. Theorists of syntax differ in exactly what they regard as a phrase; however, it is usually required to be a constituent of a sentence, in that it must include all the dependents of the units that it contains. This means that some expressions that may be called phrases in everyday language are not phrases in the technical sense. For example, in the sentence I can't put up with Alex, the words put up with (meaning \'tolerate\') may be referred to in common language as a phrase (English expressions like this are frequently called phrasal verbs\ but technically they do not form a complete phrase, since they do not include Alex, which is the complement of the preposition with.")
In Natural Language Processing, 2-word phrases are referred to as "bi-gram", and 3-word phrases are referred to as "tri-gram", and so forth. Generally, a given combination of n-words is called an "n-gram".
First, we install the ngram package (available on CRAN)
# Install package "ngram"
install.packages("ngram")
Then, we will find the most frequent two-word and three-word phrases
library(ngram)
# To find all two-word phrases in the test "text":
ng2 <- ngram(text, n = 2)
# To find all three-word phrases in the test "text":
ng3 <- ngram(text, n = 3)
Finally, we will print the objects (ngrams) using various methods as below:
print(ng, output="truncated")
print(ngram(x), output="full")
get.phrasetable(ng)
ngram::ngram_asweka(text, min=2, max=3)
We can also use Markov Chains to babble new sequences:
# if we are using ng2 (bi-gram)
lnth = 2
babble(ng = ng2, genlen = lnth)
# if we are using ng3 (tri-gram)
lnth = 3
babble(ng = ng3, genlen = lnth)

We can split the words and use table to summarize the frequency:
words <- strsplit(text, "[ ,.\\(\\)\"]")
sort(table(words, exclude = ""), decreasing = T)

Simplest?
require(quanteda)
# bi-grams
topfeatures(dfm(text, ngrams = 2, verbose = FALSE))
## of_the a_phrase the_sentence may_be as_a in_the in_common phrase_is
## 5 4 4 3 3 3 2 2
## is_usually group_of
## 2 2
# for tri-grams
topfeatures(dfm(text, ngrams = 3, verbose = FALSE))
## a_phrase_is group_of_words of_a_sentence of_the_sentence for_example_in example_in_the
## 2 2 2 2 2 2
## in_the_sentence an_orange_bird orange_bird_with bird_with_a
# 2 2 2 2

Here's a simple base R approach for the 5 most frequent words:
head(sort(table(strsplit(gsub("[[:punct:]]", "", text), " ")), decreasing = TRUE), 5)
# a the of in phrase
# 21 18 12 10 8
What it returns is an integer vector with the frequency count and the names of the vector correspond to the words that were counted.
gsub("[[:punct:]]", "", text) to remove punctuation since you don't want to count that, I guess
strsplit(gsub("[[:punct:]]", "", text), " ") to split the string on spaces
table() to count unique elements' frequency
sort(..., decreasing = TRUE) to sort them in decreasing order
head(..., 5) to select only the top 5 most frequent words

Technique for finding bad data in read.csv in R

I am reading in a file of data that looks like this:
userId, fullName,email,password,activated,registrationDate,locale,notifyOnUpdates,lastSyncTime,plan_id,plan_period_months,plan_price,plan_exp_date,plan_is_trial,plan_is_trial_used,q_hear,q_occupation,pp_subid,pp_payments,pp_since,pp_cancelled,apikey
"2","John Smith,"john.smith#gmail.com","a","1","2004-07-23 14:19:32","en_US","1","2011-04-07 07:29:17","3",\N,\N,\N,"0","1",\N,\N,\N,\N,\N,\N,"d7734dce-4ae2-102a-8951-0040ca38ff83"
but the actual file as around 20000 records. I use the following R code to read it in:
user = read.csv("~/Desktop/dbdump/users.txt", na.strings = "\\N", quote="")
And the reason I have quote="" is because without it the import stops prematurely. I end up with a total of 9569 observations. Why I don't understand why exactly the quote="" overcomes this problem, it seems to do so.
Except that it introduces other problems that I have to 'fix'. The first one I saw is that the dates end up being strings which include the quotes, which don't want to convert to actual dates when I use to.Date() on them.
Now I could fix the strings and hack my way through. But better to know more about what I am doing. Can someone explain:
Why does the quote="" fix the 'bad data'
What is a best-practice technique to figure out what is causing the read.csv to stop early? (If I just look at the input data at +/- the indicated row, I don't see anything amiss).
Here are the lines 'near' the 'problem'. I don't see the damage do you?
"16888","user1","user1#gmail.com","TeilS12","1","2008-01-19 08:47:45","en_US","0","2008-02-23 16:51:53","1",\N,\N,\N,"0","0","article","student",\N,\N,\N,\N,"ad949a8e-17ed-102b-9237-0040ca390025"
"16889","user2","user2#gmail.com","Gaspar","1","2008-01-19 10:34:11","en_US","1",\N,"1",\N,\N,\N,"0","0","email","journalist",\N,\N,\N,\N,"8b90f63a-17fc-102b-9237-0040ca390025"
"16890","user3","user3#gmail.com","boomblaadje","1","2008-01-19 14:36:54","en_US","0",\N,"1",\N,\N,\N,"0","0","article","student",\N,\N,\N,\N,"73f31f4a-181e-102b-9237-0040ca390025"
"16891","user4","user4#gmail.com","mytyty","1","2008-01-19 15:10:45","en_US","1","2008-01-19 15:16:45","1",\N,\N,\N,"0","0","google-ad","student",\N,\N,\N,\N,"2e48e308-1823-102b-9237-0040ca390025"
"16892","user5","user5#gmail.com","08091969","1","2008-01-19 15:12:50","en_US","1",\N,"1",\N,\N,\N,"0","0","dont","dont",\N,\N,\N,\N,"79051bc8-1823-102b-9237-0040ca390025"
* Update *
It's more tricky. Even though the total number of rows imported is 9569, if I look at the last few rows they correspond to the last few rows of data. Therefore I surmise that something happened during the import to cause a lot of rows to be skipped. In fact 15914 - 9569 = 6345 records. When I have the quote="" in there I get 15914.
So my question can be modified: Is there a way to get read.csv to report about rows it decides not to import?
* UPDATE 2 *
#Dwin, I had to remove na.strings="\N" because the count.fields function doesn't permit it. With that, I get this output which looks interesting but I don't understand it.
3 4 22 23 24
1 83 15466 178 4
Your second command produces a lots of data (and stops when max.print is reached.) But the first row is this:
[1] 2 4 2 3 5 3 3 3 5 3 3 3 2 3 4 2 3 2 2 3 2 2 4 2 4 3 5 4 3 4 3 3 3 3 3 2 4
Which I don't understand if the output is supposed to show how many fields there are in each record of input. Clearly the first lines all have more than 2,4,2 etc fields... Feel like I am getting closer, but still confused!

The count.fields function can be very useful in identifying where to look for malformed data.
This gives a tabulation of fields per line ignores quoting, possibly a problem if there are embedded commas:
table( count.fields("~/Desktop/dbdump/users.txt", quote="", sep=",") )
This give a tabulation ignoring both quotes and "#"(octothorpe) as a comment character:
table( count.fields("~/Desktop/dbdump/users.txt", quote="", comment.char="") )
Atfer seeing what you report for the first tabulation..... most of which were as desired ... You can get a list of the line positions with non-22 values (using the comma and non-quote settings):
which( count.fields("~/Desktop/dbdump/users.txt", quote="", sep=",") != 22)
Sometimes the problem can be solved with fill=TRUE if the only difficulty is missing commas at the ends of lines.

One problem I have spotted (thanks to data.table) is the missing quote (") after John Smith. Could this be a problem also for other lines you have?
If I add the "missing" quote after John Smith, it reads fine.
I saved this data to data.txt:
userId, fullName,email,password,activated,registrationDate,locale,notifyOnUpdates,lastSyncTime,plan_id,plan_period_months,plan_price,plan_exp_date,plan_is_trial,plan_is_trial_used,q_hear,q_occupation,pp_subid,pp_payments,pp_since,pp_cancelled,apikey
"2","John Smith","john.smith#gmail.com","a","1","2004-07-23 14:19:32","en_US","1","2011-04-07 07:29:17","3",\N,\N,\N,"0","1",\N,\N,\N,\N,\N,\N,"d7734dce-4ae2-102a-8951-0040ca38ff83"
"16888","user1","user1#gmail.com","TeilS12","1","2008-01-19 08:47:45","en_US","0","2008-02-23 16:51:53","1",\N,\N,\N,"0","0","article","student",\N,\N,\N,\N,"ad949a8e-17ed-102b-9237-0040ca390025"
"16889","user2","user2#gmail.com","Gaspar","1","2008-01-19 10:34:11","en_US","1",\N,"1",\N,\N,\N,"0","0","email","journalist",\N,\N,\N,\N,"8b90f63a-17fc-102b-9237-0040ca390025"
"16890","user3","user3#gmail.com","boomblaadje","1","2008-01-19 14:36:54","en_US","0",\N,"1",\N,\N,\N,"0","0","article","student",\N,\N,\N,\N,"73f31f4a-181e-102b-9237-0040ca390025"
"16891","user4","user4#gmail.com","mytyty","1","2008-01-19 15:10:45","en_US","1","2008-01-19 15:16:45","1",\N,\N,\N,"0","0","google-ad","student",\N,\N,\N,\N,"2e48e308-1823-102b-9237-0040ca390025"
"16892","user5","user5#gmail.com","08091969","1","2008-01-19 15:12:50","en_US","1",\N,"1",\N,\N,\N,"0","0","dont","dont",\N,\N,\N,\N,"79051bc8-1823-102b-9237-0040ca390025"
And this is a code. Both fread and read.csv works fine.
require(data.table)
dat1 <- fread("data.txt", header = T, na.strings = "\\N")
dat1
dat2 <- read.csv("data.txt", header = T, na.strings = "\\N")
dat2

Creating a vector from a file in R

I am new to R and my question should be trivial. I need to create a word cloud from a txt file containing the words and their occurrence number. For that purposes I am using the snippets package.
As it can be seen at the bottom of the link, first I have to create a vector (is that right that words is a vector?) like bellow.
> words <- c(apple=10, pie=14, orange=5, fruit=4)
My problem is to do the same thing but create the vector from a file which would contain words and their occurrence number. I would be very happy if you could give me some hints.
Moreover, to understand the format of the file to be inserted I write the vector words to a file.
> write(words, file="words.txt")
However, the file words.txt contains only the values but not the names(apple, pie etc.).
$ cat words.txt
10 14 5 4
Thanks.

words is a named vector, the distinction is important in the context of the cloud() function if I read the help correctly.
Write the data out correctly to a file:
write.table(words, file = "words.txt")
Create your word occurrence file like the txt file created. When you read it back in to R, you need to do a little manipulation:
> newWords <- read.table("words.txt", header = TRUE)
> newWords
x
apple 10
pie 14
orange 5
fruit 4
> words <- newWords[,1]
> names(words) <- rownames(newWords)
> words
apple pie orange fruit
10 14 5 4
What we are doing here is reading the file into newWords, the subsetting it to take the one and only column (variable), which we store in words. The last step is to take the row names from the file read in and apply them as the "names" on the words vector. We do the last step using the names() function.

Yes, 'vector' is the proper term.
EDIT:
A better method than write.table would be to use save() and load():
save(words. file="svwrd.rda")
load(file="svwrd.rda")
The save/load combo preserved all the structure rather than doing coercion. The write.table followed by names()<- is kind of a hassle as you can see in both Gavin's answer here and my answer on rhelp.
Initial answer:
Suggest you use as.data.frame to coerce to a dataframe an then write.table() to write to a file.
write.table(as.data.frame(words), file="savew.txt")
saved <- read.table(file="savew.txt")
saved
words
apple 10
pie 14
orange 5
fruit 4

Resources