Using ggplot to plot number of TRUE statements from a df

Question

I'm trying to plot a graph where number of TRUE statement from a df column.

I have a df that looks like this

Speed   Month_1
12      67
12      114
12      155
12      44
13      77
13      165
13      114
13      177
...

And I would like to plot a bargraph where we have x = Speed and y = Number of rows that are above 100 in Month_1 column.

So for X = 12 I would have a bargraph with a Y-value of 2 and for X = 13 I would have a Y-value of 3.

Can I do this directly in ggplot, or do I have to create a new DF first?

Answer 1

You can use dplyr to filter your data frame and then plot it with ggplot.

library(tidyverse)

df <- tibble(Speed = c(12, 12, 12, 12, 13, 13, 13, 13),
             Month_1 = c(67, 114, 155, 44, 77, 165, 114, 177))

df %>% filter(Month_1 > 100) %>% 
  ggplot(aes(x = Speed)) + geom_bar()

Answer 2

Bill's answer is good but may you like this one: just find all words with test and then filter out useless ones;

const s = "Hello this is testing _testisgood _test test ilovetesting again test"
 
console.log(
    (s.match(/[^\s]*test[^\s]*/gi) || []).filter(s => s !== 'test')
)

Answer 3

The regex below will match 'test' when it either has a non-whitespace character(s) prefixing or post fixing it.

/([^\s]+test[^\s]*|[^\s]*test[^\s]+)/gi;

OR

/(\S+test\S*|\S*test\S+)/gi;

const sentence = "Hello this is testing _testisgood _test test ilovetesting again test";

regex = /([^\s]+test[^\s]*|[^\s]*test[^\s]+)/gi;

console.log(sentence.match(regex));

Answer 4

You can match just _testisgood and ilovetesting in your example by specifying one or more characters that are not whitespace before and after test, like this:

/[^\s]+test[^\s]+/gi

If you also want to match testing, then drop [^\s]+ from the beginning of the pattern.

Answer 5

should still work like this \btest\w+|\w+test\b|\w+test\B\w+

Answer 6

Since a "word" in your scenario is a chunk of one or more non-whitespace chars (and a non-whitespace char is matched with \S in regex) you can use

console.log(
  "Hello this is testing _testisgood _test test ilovetesting again test"
    .match(/\S+test\S*|\S*test\S+/gi))
// => ["testing","_testisgood", "_test", "ilovetesting"]

Here, \S+test\S*|\S*test\S+ (see this regex demo) matches either

• \S+test\S* - one or more non-whitespace chars, test and zero or more non-whitespace chars
• | - or
• \S*test\S+ - zero or more non-whitespace chars, test and one or more non-whitespace chars.

Or, you can split with any one or more whitespace chars (with .split(/\s+/)) and then filter out any chunk that is equal to test string or does not contain test string:

console.log(
  "Hello this is testing _testisgood _test test ilovetesting again test".split(/\s+/)
    .filter(x => x.toLowerCase() != "test" && x.toLowerCase().includes("test")))

Considering your expected output it may seem that you want to match any non-whitespace chunk that contains test and at least one more letter. In that case, you can use

console.log(
  "Hello this is testing _testisgood _test test ilovetesting again test"
    .match(/[^\sa-zA-Z]*[a-zA-Z]\S*test\S*|\S*test[^\sa-zA-Z]*[a-zA-Z]\S*/gi))
// => ["testing","_testisgood", "_test", "ilovetesting"]

See this regex demo. Output: ["testing", "_testisgood", "ilovetesting"]

Details:

• [^\sa-zA-Z]* - any zero or more chars other than whitespace and letters
• [a-zA-Z] - a letter
• \S*test\S* - test enclosed with zero or more non-whitespace chars
• | - or
• \S*test[^\sa-zA-Z]*[a-zA-Z]\S* - zero or more non-whitespace chars, test, any zero or more chars other than whitespace and letters and then a letter and zero or more non-whitespace chars.

Answer 7

If what you want is:

Gives all the test but I only want the test string which is surrounded by some other character

Let's do it with another approach string built-in functions, instead of Regex

function surroundedBy(string, wordInMiddle){
  const allWords = s.split(/\s+/);
  const wordsSurrounded = allWords.filter(word=>word.toLowerCase().includes(wordInMiddle) && 
                                                        !word.startsWith(wordInMiddle) && 
                                                        !word.endsWith(wordInMiddle))
  return wordsSurrounded;
} 

Test:

const s = "Hello this is testing _testisgood _test test ilovetesting again test"
console.log(surroundedBy(s,'test'))

Result: ['_testisgood', 'ilovetesting']