wept | 微信小程序多端实时运行工具 | Chat library
kandi X-RAY | wept Summary
kandi X-RAY | wept Summary
微信小程序多端实时运行工具
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of wept
wept Key Features
wept Examples and Code Snippets
Community Discussions
Trending Discussions on wept
QUESTION
I have a large dataframe consisting of tweets, and keyword dictionaries loaded as values that have words associated with morality (kw_Moral
) and emotion (kw_Emo
). In the past I have used the keyword dictionaries to subset a dataframe to get only the tweets that have one or more of the keywords present.
For example, to create a subset with only those tweets that have emotional keywords, I loaded in my keyword dictionary...
...ANSWER
Answered 2018-Dec-12 at 14:02Your requirement would seem to lend itself to a matrix type output, where, for example, the tweets are rows, and each term is a column, with the cell value being the number of occurrences. Here is a base R solution using gsub
:
QUESTION
I have a large dataframe consisting of tweets, and a keyword dictionary loaded as a list that has words and word stems associated with emotion (kw_Emo
). I need to find a way to count how many times any given word/word stem from kw_Emo
is present each tweet. In kw_Emo
, word stems are marked with an asterisk ( * ). For example, one word stem is ador*
, meaning that I need to account for the presence of adorable
, adore
, adoring
, or any pattern of letters that starts with ador…
.
From a previous Stack Overflow discussion (see previous question on my profile), I was greatly helped with the following solution, but it only counts exact character matches (Ex. only ador
, not adorable
):
Load relevant package.
library(stringr)
Identify and remove the
*
from word stems inkw_Emo
.for (x in 1:length(kw_Emo)) { if (grepl("[*]", kw_Emo[x]) == TRUE) { kw_Emo[x] <- substr(kw_Emo[x],1,nchar(kw_Emo[x])-1) }
}Create new columns, one for each word/word stem from
kw_Emo
, with default value 0.for (x in 1:length(keywords)) { dataframe[, keywords[x]] <- 0}
Split each Tweet to a vector of words, see if the keyword is equal to any, add +1 to the appropriate word/word stems' column.
for (x in 1:nrow(dataframe)) { partials <- data.frame(str_split(dataframe[x,2], " "), stringsAsFactors=FALSE) partials <- partials[partials[] != ""] for(y in 1:length(partials)) { for (z in 1:length(keywords)) { if (keywords[z] == partials[y]) { dataframe[x, keywords[z]] <- dataframe[x, keywords[z]] + 1 } } } }
Is there a way to alter this solution to account for word stems? I'm wondering if it's possible to first use a stringr pattern to replace occurrences of a word stem with the exact characters, and then use this exact match solution. For instance, something like stringr::str_replace_all(x, "ador[a-z]+", "ador")
. But I'm unsure how to do this with my large dictionary and numerous word stems. Maybe the loop removing [*]
, which essentially identifies all word stems, can be adapted somehow?
Here is a reproducible sample of my dataframe, called TestTweets
with the text to be analysed in a column called clean_text
:
dput(droplevels(head(TestTweets, 20)))
ANSWER
Answered 2019-Jan-08 at 12:17So first of all I would get rid of some of the for
loops:
QUESTION
So I'm writing a program that generates a sentence using BNF grammar. So let's say that I had this for a grammar file:
...ANSWER
Answered 2017-May-03 at 02:51 ::= '\x0A'
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install wept
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page