Hey friends I've been using chrome for the development of my websites but know I wanna switch things up a bit. I read this article on how to do it. I did this in the past for chrome and it worked. But when I paste:
[
{ “keys”: [ “ctrl+alt+v” ], “command”: “view_in_browser” },
{ “keys”: [ “ctrl+alt+f” ], “command”: “view_in_browser”, “args”: { “browser”: “firefox” } },
{ “keys”: [ “ctrl+alt+c” ], “command”: “view_in_browser”, “args”: { “browser”: “chrome” } },
{ “keys”: [ “ctrl+alt+i” ], “command”: “view_in_browser”, “args”: { “browser”: “iexplore” } },
{ “keys”: [ “ctrl+alt+s” ], “command”: “view_in_browser”, “args”: { “browser”: “safari” } }
]
in the key bindings user file I get this error?
Error trying to parse file: Expected value in Packages\User\Default
(Windows).sublime-keymap:3:4
Edit: I was told to turn the curly quotes into straight quotes. I did this and while it did fix the issue of saving the file, the error message did not show up. However I am not able to open Firefox with Ctrl + Alt + f?
The reason that this is not working for you is that your JSON is invalid; JSON only allows for straight double quotes, but the JSON you pasted above is using curly quotes:
From https://en.wikipedia.org/wiki/Quotation_mark:
'...' and "…" are known as neutral, vertical, straight, typewriter, dumb, or ASCII quotation marks. The left and right marks are identical. These are found on typical English typewriters and computer keyboards, although they are sometimes automatically converted to the other type by software.
‘…’ and “…” are known as typographic, curly, curved, book, or smart quotation marks. The beginning marks are commas raised to the top of the line and rotated 180 degrees. The ending marks are commas raised to the top of the line. Curved quotation marks are used mainly in manuscript, printing and typesetting.
As a result of this, Sublime's JSON parser (which is lenient in that it allows for extraneous trailing commas and comments that standard JSON disallows) doesn't understand the curly quotes, so it's not finding what it expects at line 3, column 4.
Replacing all of the double quote characters with straight quotes should solve the problem.
I have a list of sample names:
TW1
UD1
SS1
S17
SS23
UD12
I wish to add a hyphen in between the letters and the numbers as such:
TW-1
UD-1
SS-1
S-17
SS-23
UD-12
UD786
I tried this:
=MID(A1,1,COUNT(1*MID(A1,{1,2,3,4,5,6,7,8,9,0},1)))&"-"&SUBSTITUTE(A1,MID(A1,1,COUNT(1*MID(A1,{1,2,3,4,5,6,7,8,9,0},1))),"")
The results were not consistent. It gives the following results:
T-W1
U-D1
S-1
S1-7
SS-23
UD-12
How may I achieve the desired output?
=LEFT(A1,2-ISNUMBER(--MID(A1,2,1)))&"-"&RIGHT(A1,LEN(A1)-(2-ISNUMBER(--MID(A1,2,1))))
Is another option. Trick in this is to convert your number as a string to a number before testing if its a number. Instead of -- you could have done 1* or 0+.
This solution only works as your sample data was 1 or 2 characters before the first digit.
Slightly shorter than #Forward Ed's A:
=REPLACE(A1,2+(CODE(MID(A1,2,1))>64),,"-")
I'm trying to search across a large array of textual files in Mathematica 8 (12k+). So far, I've been able to plot the sheer numbers of times that a word appears (i.e. the word "love" appears 5,000 times across those 12k files). However, I'm running into difficulty determining the number of files in which "love" appears once - which might only be in 1,000 files, with it repeating several times in others.
I'm finding the documentation WRT FindList, streams, RecordSeparators, etc. a bit murky. Is there a way to set it up so it finds an incidence of a term once in a file and then moves onto the next?
Example of filelist:
{"89001.txt", "89002.txt", "89003.txt", "89004.txt", "89005.txt", "89006.txt", "89007.txt", "89008.txt", "89009.txt", "89010.txt", "89011.txt", "89012.txt", "89013.txt", "89014.txt", "89015.txt", "89016.txt", "89017.txt", "89018.txt", "89019.txt", "89020.txt", "89021.txt", "89022.txt", "89023.txt", "89024.txt"}
The following returns all of the lines with love across every file. Is there a way to return only the first incidence of love in each file before moving onto the next one?
FindList[filelist, "love"]
Thanks so much. This is my first post and I'm largely learning Mathematica through peer/supervisory help, online tutorials, and the documentation.
In addition to Daniel's answer, you also seem to be asking for a list of files where the word only occurs once. To do that, I'd continue to run FindList across all the files
res =FindList[filelist, "love"]
Then, reduce the results to single lines only, via
lines = Select[ res, Length[#]==1& ]
But, this doesn't eliminate the cases where there is more than one occurrence in a single line. To do that, you could use StringCount and only accept instances where it is 1, as follows
Select[ lines, StringCount[ #, RegularExpression[ "\\blove\\b" ] ] == 1& ]
The RegularExpression specifies that "love" must be a distinct word using the word boundary marker (\\b), so that words like "lovely" won't be included.
Edit: It appears that FindList when passed a list of files returns a flattened list, so you can't determine which item goes with which file. For instance, if you have 3 files, and they contain the word "love", 0, 1, and 2 times, respectively, you'd get a list that looked like
{, love, love, love }
which is clearly not useful. To overcome this, you'll have to process each file individually, and that is best done via Map (/#), as follows
res = FindList[#, "love"]& /# filelist
and the rest of the above code works as expected.
But, if you want to associate the results with a file name, you have to change it a little.
res = {#, FindList[#, "love"]}& /# filelist
lines = Select[res,
Length[ #[[2]] ] ==1 && (* <-- Note the use of [[2]] *)
StringCount[ #[[2]], RegularExpression[ "\\blove\\b" ] ] == 1&
]
which returns a list of the form
{ {filename, { "string with love in it" },
{filename, { "string with love in it" }, ...}
To extract the file names, you simply type lines[[All, 1]].
Note, in order to Select on the properties you wanted, I used Part ([[ ]]) to specify the second element in each datum, and the same goes for extracting the file names.
Help > Documentation Center > FindList item 4:
"FindList[files,text,n]
includes only the first n lines found."
So you could set n to 1.
Daniel Lichtblau
I have an Excel spreadsheet containing a list of strings. Each string is made up of several words, but the number of words in each string is different.
Using built in Excel functions (no VBA), is there a way to isolate the last word in each string?
Examples:
Are you classified as human? -> human?
Negative, I am a meat popsicle -> popsicle
Aziz! Light! -> Light!
This one is tested and does work (based on Brad's original post):
=RIGHT(A1,LEN(A1)-FIND("|",SUBSTITUTE(A1," ","|",
LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
If your original strings could contain a pipe "|" character, then replace both in the above with some other character that won't appear in your source. (I suspect Brad's original was broken because an unprintable character was removed in the translation).
Bonus: How it works (from right to left):
LEN(A1)-LEN(SUBSTITUTE(A1," ","")) – Count of spaces in the original string
SUBSTITUTE(A1," ","|", ... ) – Replaces just the final space with a |
FIND("|", ... ) – Finds the absolute position of that replaced | (that was the final space)
Right(A1,LEN(A1) - ... )) – Returns all characters after that |
EDIT: to account for the case where the source text contains no spaces, add the following to the beginning of the formula:
=IF(ISERROR(FIND(" ",A1)),A1, ... )
making the entire formula now:
=IF(ISERROR(FIND(" ",A1)),A1, RIGHT(A1,LEN(A1) - FIND("|",
SUBSTITUTE(A1," ","|",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))
Or you can use the =IF(COUNTIF(A1,"* *") syntax of the other version.
When the original string might contain a space at the last position add a trim function while counting all the spaces: Making the function the following:
=IF(ISERROR(FIND(" ",B2)),B2, RIGHT(B2,LEN(B2) - FIND("|",
SUBSTITUTE(B2," ","|",LEN(TRIM(B2))-LEN(SUBSTITUTE(B2," ",""))))))
This is the technique I've used with great success:
=TRIM(RIGHT(SUBSTITUTE(A1, " ", REPT(" ", 100)), 100))
To get the first word in a string, just change from RIGHT to LEFT
=TRIM(LEFT(SUBSTITUTE(A1, " ", REPT(" ", 100)), 100))
Also, replace A1 by the cell holding the text.
A more robust version of Jerry's answer:
=TRIM(RIGHT(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))), LEN(TRIM(A1))))
That works regardless of the length of the string, leading or trailing spaces, or whatever else and it's still pretty short and simple.
I found this on google, tested in Excel 2003 & it works for me:
=IF(COUNTIF(A1,"* *"),RIGHT(A1,LEN(A1)-LOOKUP(LEN(A1),FIND(" ",A1,ROW(INDEX($A:$A,1,1):INDEX($A:$A,LEN(A1),1))))),A1)
[edit] I don't have enough rep to comment, so this seems the best place...BradC's answer also doesn't work with trailing spaces or empty cells...
[2nd edit] actually, it doesn't work for single words either...
=RIGHT(TRIM(A1),LEN(TRIM(A1))-FIND(CHAR(7),SUBSTITUTE(" "&TRIM(A1)," ",CHAR(7),
LEN(TRIM(A1))-LEN(SUBSTITUTE(" "&TRIM(A1)," ",""))+1))+1)
This is very robust--it works for sentences with no spaces, leading/trailing spaces, multiple spaces, multiple leading/trailing spaces... and I used char(7) for the delimiter rather than the vertical bar "|" just in case that is a desired text item.
This is very clean and compact, and works well.
{=RIGHT(A1,LEN(A1)-MAX(IF(MID(A1,ROW(1:999),1)=" ",ROW(1:999),0)))}
It does not error trap for no spaces or one word, but that's easy to add.
Edit:
This handles trailing spaces, single word, and empty cell scenarios. I have not found a way to break it.
{=RIGHT(TRIM(A1),LEN(TRIM(A1))-MAX(IF(MID(TRIM(A1),ROW($1:$999),1)=" ",ROW($1:$999),0)))}
=RIGHT(A1,LEN(A1)-FIND("`*`",SUBSTITUTE(A1," ","`*`",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
New answer 9/28/2022
Considering the new excel function: TEXTAFTER (check availability) you can achieve it with a simple formula:
=TEXTAFTER(A1," ", -1)
To add to Jerry and Joe's answers, if you're wanting to find the text BEFORE the last word you can use:
=TRIM(LEFT(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))), LEN(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))))-LEN(TRIM(A1))))
With 'My little cat' in A1 would result in 'My little' (where Joe and Jerry's would give 'cat'
In the same way that Jerry and Joe isolate the last word, this then just gets everything to the left of that (then trims it back)
Copy into a column, select that column and HOME > Editing > Find & Select, Replace:
Replace All.
There is a space after the asterisk.
Imagine the string could be reversed. Then it is really easy. Instead of working on the string:
"My little cat" (1)
you work with
"tac elttil yM" (2)
With =LEFT(A1;FIND(" ";A1)-1) in A2 you get "My" with (1) and "tac" with (2), which is reversed "cat", the last word in (1).
There are a few VBAs around to reverse a string. I prefer the public VBA function ReverseString.
Install the above as described. Then with your string in A1, e.g., "My little cat" and this function in A2:
=ReverseString(LEFT(ReverseString(A1);IF(ISERROR(FIND(" ";A1));
LEN(A1);(FIND(" ";ReverseString(A1))-1))))
you'll see "cat" in A2.
The method above assumes that words are separated by blanks. The IF clause is for cells containing single words = no blanks in cell. Note: TRIM and CLEAN the original string are useful as well. In principle it reverses the whole string from A1 and simply finds the first blank in the reversed string which is next to the last (reversed) word (i.e., "tac "). LEFT picks this word and another string reversal reconstitutes the original order of the word (" cat"). The -1 at the end of the FIND statement removes the blank.
The idea is that it is easy to extract the first(!) word in a string with LEFT and FINDing the first blank. However, for the last(!) word the RIGHT function is the wrong choice when you try to do that because unfortunately FIND does not have a flag for the direction you want to analyse your string.
Therefore the whole string is simply reversed. LEFT and FIND work as normal but the extracted string is reversed. But his is no big deal once you know how to reverse a string. The first ReverseString statement in the formula does this job.
=LEFT(A1,FIND(IF(
ISERROR(
FIND("_",A1)
),A1,RIGHT(A1,
LEN(A1)-FIND("~",
SUBSTITUTE(A1,"_","~",
LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))
)
)
)
),A1,1)-2)
I translated to PT-BR, as I needed this as well.
(Please note that I've changed the space to \ because I needed the filename only of path strings.)
=SE(ÉERRO(PROCURAR("\",A1)),A1,DIREITA(A1,NÚM.CARACT(A1)-PROCURAR("|", SUBSTITUIR(A1,"\","|",NÚM.CARACT(A1)-NÚM.CARACT(SUBSTITUIR(A1,"\",""))))))
Another way to achieve this is as below
=IF(ISERROR(TRIM(MID(TRIM(D14),SEARCH("|",SUBSTITUTE(TRIM(D14)," ","|",LEN(TRIM(D14))-LEN(SUBSTITUTE(TRIM(D14)," ","")))),LEN(TRIM(D14))))),TRIM(D14),TRIM(MID(TRIM(D14),SEARCH("|",SUBSTITUTE(TRIM(D14)," ","|",LEN(TRIM(D14))-LEN(SUBSTITUTE(TRIM(D14)," ","")))),LEN(TRIM(D14)))))
You can achieve this also by reversing the string and finding the first space
=MID(C3,2+LEN(C3)-SEARCH(" ",CONCAT(MID(C3,SEQUENCE(LEN(C3),,LEN(C3),-1),1))),LEN(A1))
Reverse the string
CONCAT(MID(C3,SEQUENCE(LEN(C3),,LEN(C3),-1),1))
Find the first space in the reversed string
SEARCH(" ",...
Take the position of the space found in the reversed string off the length of the string and return that portion
=MID(C3,2+LEN(C3)-SEARCH...
I also had a task like this and when I was done, using the above method, a new method occured to me: Why don't you do this:
Reverse the string ("string one" becomes "eno gnirts").
Use the good old Find (which is hardcoded for left-to-right).
Reverse it into readable string again.
How does this sound?