i am doing "find and repleace button" for my application. I am using gtk and ruby. And i can find that how many word, if there is. Also i want to get selection word that searched word, and i should mark them. My some code:
def search(ent, txtvu)
start = txtvu.buffer.start_iter
first, last = start.forward_search(ent.text, Gtk::TextIter::SEARCH_TEXT_ONLY, nil)
count = 0
while (first)
mark = start.buffer.create_mark(nil, first, false)
txtvu.scroll_mark_onscreen(mark)
txtvu.buffer.delete_mark(mark)
txtvu.buffer.select_range(first, last)
start.forward_char
first, last = start.forward_search(ent.text, Gtk::TextIter::SEARCH_TEXT_ONLY, nil)
start = first
count += 1
end
count says me how many words involve My code does't work. :( Why? I want to mark all searched words.
If I understand you correctly, you want to highlight all found words, not just one. In that case, select_range is not the function to call, because it will change the selection to the current word, and GtkTextView selection is single and contiguous.
Instead, create a highlight tag and apply it to all searches. For example:
# create the "highlight" tag (run this only once)
textvu.buffer.create_tag("highlight", {background => "yellow"})
# ... later, in the loop:
textvu.buffer.apply_tag("highlight", first, last)
Your matches will all appear highlighted.
Related
I'm trying to compare Items in "projectEstimate!D2:D & projectEstimate!E2:E" to Items in 'itemsAssociations!C3:C & itemsAssociations!D3:D" for matches.
If there is a match, confirm that the associated item (itemsAssociations!G3:G & itemsAssociations!H3:H) is not already listed in "projectEstimate".
If it is not listed, print that item. If it IS listed, do nothing.
I put together the following code which seems like it should work, but the item prints whether it's present on projectEstimate or not.
=ArrayFormula({itemsAssociations!I2:J2;FILTER(itemsAssociations!I3:J,
{projectEstimate!D2:D&IF(LEN(projectEstimate!E2:E),projectEstimate!E2:E,)=itemsAssociations!C3:C&IF(LEN(itemsAssociations!D3:D),itemsAssociations!D3:D,)},
{projectEstimate!D2:D&IF(LEN(projectEstimate!E2:E),projectEstimate!E2:E,)<>itemsAssociations!G3:G&IF(LEN(itemsAssociations!H3:H),itemsAssociations!H3:H,)}
)})
I also tried this QUERY, but not sure how to include the entire ranges
=QUERY(itemsAssociations!C3:J,"SELECT I,J WHERE C = '"&projectEstimate!D2:D&"' AND D = '"&projectEstimate!E2:E&"' AND J != '"&projectEstimate!D2:D&"'",0)
This is close, but the opposite result:
=FILTER(projectEstimate!D2:D,COUNTIF(FILTER(itemsAssociations!J3:J,COUNTIF(itemsAssociations!C3:C&itemsAssociations!D3:D,itemsAssociations!C3:C&itemsAssociations!D3:D)),projectEstimate!D2:D))
My sheet
it would be something amongst theses lines:
=ARRAYFORMULA(REGEXREPLACE(QUERY(FILTER(projectEstimate!D2:D&":"&projectEstimate!E2:E,
NOT(COUNTIF(itemsAssociations!C3:C&":"&itemsAssociations!D2:D,projectEstimate!D2:D&":"&projectEstimate!E2:E))),
"where Col1 is not null", 0), ":$", ))
This is what I came up with...
=IFERROR(FILTER(FILTER(itemsAssociations!J3:J,NOT(COUNTIF(projectEstimate!D3:D,itemsAssociations!C3:C))),NOT(COUNTIF(FILTER(projectEstimate!D2:E,NOT(COUNTIF(itemsAssociations!C3:C&itemsAssociations!D3:D,projectEstimate!D2:D&projectEstimate!E2:E))),FILTER(itemsAssociations!J3:J,NOT(COUNTIF(projectEstimate!D3:D,itemsAssociations!C3:C)))))),"No suggested items!")
A page loads 35.000 elements, which only the first 10 are of interest to me. Returning all elements makes the scraping extremely slow.
I only succeeded in either returning the first element with:
driver.find_element_by
Or returning all, 35.000 elements, with:
driver.find_elements_by
Anyone knows a way to return x amount of elements found?
Selenium does not provide a facility that allows returning only a slice of the .find_elements... calls. A general solution if you want to optimize things so that you do not need to have Selenium return every single element is perform the slice operation on the browser side, in JavaScript. I present this solution in this answer here. If you want to use XPath for selecting the DOM nodes, you could adapt the answer here to that, or you could use the method in another answer I've submitted.
from selenium import webdriver
driver = webdriver.Chrome()
driver.get("http://www.example.com")
# We add 35000 paragraphs with class `test` to the page so that we can
# later show how to get the first 10 paragraphs of this class. Each
# paragraph is uniquely numbered.
driver.execute_script("""
var html = [];
for (var i = 0; i < 35000; ++i) {
html.push("<p class='test'>"+ i + "</p>");
}
document.body.innerHTML += html.join("");
""")
elements = driver.execute_script("""
return Array.prototype.slice.call(document.querySelectorAll("p.test"), 0, 10);
""")
# Verify that we got the first 10 elements by outputting the text they
# contain to the console. The loop here is for illustration purposes
# to show that the `elements` array contains what we want. In real
# code, if I wanted to process the text of the first 10 elements, I'd
# do what I show next.
for element in elements:
print element.text
# A better way to get the text of the first 10 elements. This results
# in 1 round-trip between this script and the browser. The loop above
# would take 10 round-trips.
print driver.execute_script("""
return Array.prototype.slice.call(document.querySelectorAll("p.test"), 0, 10)
.map(function (x) { return x.textContent; });;
""")
driver.quit()
The Array.prototype.slice.call rigmarole is needed because what document.querySelectorAll returns looks like an Array but is not actually an Array object. (It is a NodeList.) So it does not have a .slice method but you can pass it to Array's slice method.
Here is a significantly different approach presented as a different answer because some people will prefer this one to the other one I gave, or the other one to this one.
This one relies on using XPath to slice the results:
from selenium import webdriver
driver = webdriver.Chrome()
driver.get("http://www.example.com")
# We add 35000 paragraphs with class `test` to the page so that we can
# later show how to get the first 10 paragraphs of this class. Each
# paragraph is uniquely numbered. These paragraphs are put into
# individual `div` to make sure they are not siblings of one
# another. (This prevents offering a naive XPath expression that would
# work only if they *are* siblings.)
driver.execute_script("""
var html = [];
for (var i = 0; i < 35000; ++i) {
html.push("<div><p class='test'>"+ i + "</p></div>");
}
document.body.innerHTML += html.join("");
""")
elements = driver.find_elements_by_xpath(
"(//p[#class='test'])[position() < 11]")
for element in elements:
print element.text
driver.quit()
Note that XPath uses 1-based indexes so < 11 is indeed the proper expression. The parentheses around the first part of the expression are absolutely necessary. With these parentheses, the [position() < 11] test checks the position each node has in the nodeset which is the result of the expression in parentheses. Without them, the position test would check the position of the nodes relative to their parents nodes, which would match all nodes because all <p> are at the first position in their respective <div>. (This is why I've added those <div> elements above: to show this problem.)
I would use this solution if I were already using XPath for my selection. Otherwise, if I were doing a search by CSS selector or by id I would not convert it to XPath only to perform the slice. I'd use the other method I've shown.
I'm trying to scrape a tracklist from a website. My relevant code is:
page.css('ol').each do |line|
subarray = line.text.strip.split(" - ")
end
This makes the array take the first artist into the first index (as I want), but adds the track and the artist of track two into the second index like this:
subarray[0] = Rick Wilhite
subarray[1] = Magic Water [Still Music]
Edward
subarray[2] = Into A Better Future [Giegling]
Kassem Mosse
subarray[3] = Zolarem [Mikrodisko Recordings]
After Hours
I included the nested tag so my code reads:
page.css('ol li').each do |line|
subarray = line.text.strip.split(" - ")
end
but this only seems to leave subarray[0] displaying "Klara Lewis" and subarray[1] displaying "Shine [Editions Mego]", which is the last track on the tracklist. All other index values are blank.
A further complication is that I would like to remove the record label from what will end up being the track value. I believe the correct regular expression is \[[\d\D]*?\], but I'm under the impression that this needs to be applied before the data goes into the array to avoid complications involved in iterating over arrays. I tried passing it as a second delimiter to split (along with ' - ') which didn't work, and I also attempted to test it by changing my code to:
page.css('ol').each do |line|
subarray = line.text.strip.split("\[[\d\D]*?\]")
end
but that also appears not to work. Can anyone help me on this or give me the right pointers?
Here's what's happening:
page.css('ol') gives you the entire <ol> with every one of the <li> tags:
<ol>
<li>Rick Wilhite...</li>
<li>Edward...</li>
...
<li>Klara Lewis...</li>
</ol>
When that one big chunk enters the .each loop, you're only running through the loop once. So when you apply the .split(" - ") method, subarray will be filled once with all the text separated by -.
On the other hand, page.css('ol li') gives you each individual <li>, like this:
<li>Rick Wilhite...</li>
<li>Edward...</li>
...
<li>Klara Lewis...</li>
This time, you're running through the loop 17 times, once for each <li> tag. The first time through, .split(" - ") is applied to the text and stored in the subarray variable. The problem is that the next time through the loop, subarray is overwritten with the split text of the second <li>. So after the final time through, the only contents of the subarray variable is the split text of the final <li>: "Klara Lewis" and "Shine [Editions Mego]".
I think you've gotten the general idea of how to scrape from a website, but I recommend building your script more incrementally so you understand exactly what you're doing in each step. For example, use puts to check what page.css('ol') gives you and how it differs from page.css('ol li'). What happens when it goes through a loop? What do you get when you apply .split()? Building more slowly and exploring around to make sure you understand what you're doing will help you avoid hitting dead ends. Hope that helps!
My question is, how can I search through an array and replace the string at the current index of the search without knowing what the indexed array string contains?
The code below will search through an ajax file hosted on the internet, it will find the inventory, go through each weapon in my inventory, adding the ID to a string (so I can check if that weapon has been checked before). Then it will add another value after that of the amount of times it occurs in the inventory, then after I have check all weapon in the inventory, it will go through the all of the IDs added to the string and display them along with the number (amount of occurrences). This is so I know how many of each weapon I have.
This is an example of what I have:
strList = ""
inventory.each do |inv|
amount = 1
exists = false
ids = strList.split(',')
ids.each do |ind|
if (inv['id'] == ind.split('/').first) then
exists = true
amount = ind.split('/').first.to_i
amount += 1
ind = "#{inv['id']}/#{amount.to_s}" # This doesn't seem work as expected.
end
end
if (exists == true) then
ids.push("#{inv['id']}/#{amount.to_s}")
strList = ids.join(",")
end
end
strList.split(",").each do |item|
puts "#{item.split('/').first} (#{item.split('/').last})"
end
Here is an idea of what code I expected (pseudo-code):
inventory = get_inventory()
drawn_inv = ""
loop.inventory do |inv|
if (inv['id'].occurred_before?)
inv['id'].count += 1
end
end loop
loop.inventory do |inv|
drawn_inv.add(inv['id'] + "/" + inv['id'].count)
end loop
loop.drawn_inv do |inv|
puts "#{inv}"
end loop
Any help on how to replace that line is appreciated!
EDIT: Sorry for not requiring more information on my code. I skipped the less important part at the bottom of the code and displayed commented code instead of actual code, I'll add that now.
EDIT #2: I'll update my description of what it does and what I'm expecting as a result.
EDIT #3: Added pseudo-code.
Thanks in advance,
SteTrezla
You want #each_with_index: http://ruby-doc.org/core-2.2.0/Enumerable.html#method-i-each_with_index
You may also want to look at #gsub since it takes a block. You may not need to split this string into an array at all. Basically something like strList.gsub(...){ |match| #...your block }
For a project that I am working on for school, one of the parts of the project asks us to take a collection of all the Federalist papers and run it through a program that essentially splits up the text and writes new files (per different Federalist paper).
The logic I decided to go with is to run a search, and every time the search is positive for "Federalist No." it would save into a new file everything until the next "Federalist No".
This is the algorithm that I have so far:
file_name = "Federalist"
section_number = "1"
new_text = File.open(file_name + section_number, 'w')
i = 0
n= 1
while i < l.length
if (l[i]!= "federalist") and (l[i+1]!= "No")
new_text.puts l[i]
i = i + i
else
new_text.close
section_number = (section_number.to_i +1).to_s
new_text = File.open(file_name + section_number, "w")
new_text.puts(l[i])
new_text.puts(l[i+1])
i=i+2
end
end
After debugging the code as much as I could (I am a beginner at Ruby), the problem that I run into now is that because the while function always holds true, it never proceeds to the else command.
In terms of going about this in a different way, my TA suggested the following:
Put the entire text in one string by looping through the array(l) and adding each line to the one big string each time.
Split the string using the split method and the key word "FEDERALIST No." This will create an array with each element being one section of the text:
arrayName = bigString.split("FEDERALIST No.")
You can then loop through this new array to create files for each element using a similar method you use in your program.
But as simple as it may sound, I'm having an extremely difficult time putting even that code together.
i = i + i
i starts at 0, and 0 gets added to it, which gives 0, which will always be less than l, whatever that value is/means.
Since this is a school assignment, I hesitate to give you a straight-up answer. That's really not what SO is for, and I'm glad that you haven't solicited a full solution either.
So I'll direct you to some useful methods in Ruby instead that could help.
In Array: .join, .each or .map
In String: .split
Fyi, your TA's suggestion is far simpler than the algorithm you've decided to embark on... although technically, it is not wrong. Merely more complex.