I'm trying to create a CSV file (Excel or LibreOffice) which contains a formula that uses multiple arguments separated by a comma. The comma is interpreted as a field separator. Is it possible to escape the comma somehow so formula requiring commas can be used?
This works as expected:
=Sum(A1:A10)
This doesn't read correctly due to use of comma in formula:
=Confidence.Norm(.01, Stdev.p(A1:A10), 10)
Imagined solutions that don't work:
=Confidence.Norm(.01 \, Stdev.p(A1:A10) \, 10)
=Confidence.Norm(.01 ',' Stdev.p(A1:A10) ',' 10)
=+"Confidence.Norm(.01 \, Stdev.p(A1:A10) \, 10)"
Try this:
"=Confidence.Norm(.01, Stdev.p(A1:A10), 10)"
Related
In vim, in a Windows machine (with no access to "unix"-like commands such command column) I want to reformat this code to make it more readable:
COLUMN KEY_ID FORMAT 9999999999
COLUMN VALUE_1 FORMAT 99
COLUMN VALUE_2 FORMAT 99
COLUMN VALUE_3 FORMAT 999
COLUMN VALUE_4 FORMAT 999
And I want to have this using as less commands as possible:
COLUMN KEY_ID FORMAT 9999999999
COLUMN VALUE_1 FORMAT 99
COLUMN VALUE_2 FORMAT 99
COLUMN VALUE_3 FORMAT 999
COLUMN VALUE_4 FORMAT 999
Note this is just an excerpt, as there many more lines in which I must do the same.
You could use the following command:
:%s/\w\zs\s*\zeFORMAT/^I
The pattern will match the whitespaces between FORMAT and the end of the previous word and replace it by a tab:
\w Any 'word' character
\zs Start the matching
\s* Any number of whitespace
\ze End the matching
FORMAT The actual word format
\zs and \ze allow to apply the substitution only on the whitespaces see: :h /\zs and :h /\ze
Note that ^I should be inserted with ctrl+vtab
The tabular plugin recommended by #SatoKatsura would be a good way to do it too.
You can also generalize that. Let's say you have the following file:
COLUMN KEY_ID FORMAT 9999999999
COLUMN VALUE_1 FOO 99
COLUMN VALUE_2 BAR 99
You could use this command:
:%s/^\(\w*\s\)\{1}\w*\zs\s*\ze/
Were the pattern can be detailed like that:
^ Match the beginning of the line
\(\w*\s\)\{1} One occurrence of the pattern \w*\s i.e. one column
\w* Another column
\zs\s*\ze The whitespaces after the previous column
You could change the value of \{1} to apply the command on the next columns.
EDIT to answer #aturegano comment, here is a way to align the column to another one:
%s/^\(\w*\s\)\{1}\w*\zs\s*\ze/\=repeat(' ', 30-matchstrpos(getline('.'), submatch(0))[1])
The idea is still to match the whitespaces which must be aligned, on the second part of the substitution command we use a sub-replace-expression (See :h sub-replace-expression).
This allows us to use a command from the substitution part, which can be explained like this:
\= Interpret the next characters as a command
repeat(' ', XX) Replace the match with XX whitespaces
XX is decomposed like this:
30- 30 less the next expression
matchstrpos()[1] Returns the columns where the second argument appears in the first one
getline('.') The current line (i.e. the one containing the match
submatch(0) The matched string
[1] Necessary since matchstrpos() returns a list:
[matchedString, StartPosition, EndPosition]
and we are looking for the second value.
You then simply have to replace 30 by the column where you want to move your next column.
See :h matchstrpos(), :h getline() and :h submatch()
For alignment, there are three well-known plugins:
the venerable Align - Help folks to align text, eqns, declarations, tables, etc
the modern tabular
the contender vim-easy-align
Posting an answer as requested:
:g/^COLUMN / s/.*/\=call('printf', ['%s %-30s %s %s'] + split(submatch(0)))/
Explanation:
g/^COLUMN / - apply the following command to lines matching /^COLUMN / (cf. :h :global)
\= - replace with the result of evaluating an expression, rather than with a fixed string (cf. :h s/\=)
submatch(0) - the line being matched
split(...) - split line into words
printf(...) - format the line
call(...) - we'd like to have printf('%s %-30s %s %s', list), but printf() doesn't take "real" lists as arguments, so we have to unfold the list with a call(...) (cf. :h call()).
Yet another solution:
:%s/ \{2,}/ /g
This solution is not perfect because the result will have an extra single space on the first line. To fix this problem:
:%s/\%>15c \{2,}/ /g
Explanation of pattern:
%>15c\s\{2,}
%>15c Matches only after column 15
\s\{2,} Matches two or more white spaces
I have a comma delimited file (CSV file) test.csv as shown below.
FHEAD,1,2,3,,,,,,
FDEP,2,3,,,,,,,,
FCLS,3,,,4-5,,,,,,,
FDETL,4,5,6,7,8,
FTAIL,5,67,,,,,,
I wanted to remove the empty columns only from 2nd and 3rd row of the file i.e. were ever the records starts with FDEP and FCLS only in those rows I wanted to remove the empty columns (,,).
after removing the empty columns the same file test.csv should look like
FHEAD,1,2,3,,,,,,
FDEP,2,3
FCLS,3,4-5
FDETL,4,5,6,7,8,
FTAIL,5,67,,,,,,
How can I do this in Unix???
Here's one way to do it, using sed:
sed '/^F\(DEP\|CLS\),/ { s/,\{2,\}/,/g; s/,$// }'
We use a range of /^F\(DEP\|CLS\),/, i.e. the following command will only process lines matching ^F\(DEP\|CLS\),. This regex matches beginning-of-string, followed by F, followed by either DEP or CLS, followed by ,. In other words, we look for lines starting with FDEP, or FCLS,.
Having found such a line, we first substitute (s command) all runs (g flag, match as many times as possible) of 2 or more (\{2,\}) commas (,) in a row by a single ,. This squeezes ,,, down to a single ,.
Second, we substitute , at end-of-string by nothing. This gets rid of any trailing comma.
Now I use this code to read data from standard input:
print =: 1!:2&2
read =: 1!:1[3
in =. (read-.LR)-.CR
But it returns just a sequence of numbers, e.g. input:
2
3
4
5
Output:
2345
Number of numbers is unknown, but each is in the separate line
When reading with (1!:1) you read a stream of characters. You have to manipulate the stream to get your desired input.
For example. If you want to enter a list of line separated integers, you would read the list, then split it by LF, remove LF and then convert to integer. You can achieve the first two steps using cut (;._2) and the conversion using do (".):
in =: ".;._2 (1!:1) 3
If you want to enter a list of space separated integers, you would just use do, the splitting would be implied by the spaces:
in =: ". LF -.~ (1!:1) 3
trailing LF (if present) has to be removed before applying ". because do can't convert special characters.
I have a huge data set of entries like these:
(21, 2, '23.5R25 ETADT', 'description, with a comma'),
(22, 1, '26.5R25 ETADT', 'Description without a comma'),
(23, 5, '20.5R20.5', 'Another description with ; semicolumn'),
I'm trying to replace every comma in the list with a tab. Excluding the commas within the single quotes. Also excluding the ending commas.
So the examples entries should become:
(21[TAB]2[TAB]'23.5R25 ETADT'[TAB]'description, with a comma'),
(22[TAB]1[TAB]'26.5R25 ETADT'[TAB]'Description without a comma'),
(23[TAB]5[TAB]'20.5R20.5'[TAB]'Another description with ; semicolumn'),
I've got like 6000 rows of data like this. The tabs allow me to tell Excel to import the elements of these entries into different columns.
The Regex I've tried was this: [ ]*,[ ]*
But this Regex selects all the commas, even the ones within the single quotes.
It looks as though each of your lines has 4 elements within parenthesis. And it looks like only the last 2 elements use single quotes. If those assumptions can be made, I've tested the following in Notepad++:
"Find what :" ^\(([^,]*),\s*([^,]*),\s*'([^']*)'\s*,\s*
"Replace with :" \(\1\t\2\t'\3'\t
EDIT:
The search regex is dependent upon the 4 column model with only the last two elements having single quotes. Visually this is how it works:
^\(: Finds an opening parenthesis
([^,]*): Captures non-comma characters which will be all of element 1
,\s*: Matches a comma and any trailing spaces
([^,]*): Captures non-comma characters which will be all of element 2
,\s*: Matches a comma and any trailing spaces
'([^']*)': Captures the string in single quotes which will be all of element 3
\s*,\s*: Matches a comma and all surrounding spaces
Ignore the rest of the string, there are no more commas to be replaced we just want to replace parts of the line we just read in
I'm searching to empty out multiple columns in VIM
(not to delete but to put spaces inside).
This is my search command:
/\%2c\|\%4c\|\%>5c\%<9c
(column: 2,4,6-8)
How can I empty out these columns in vim?
:%s/\%2c\|\%4c\|\%>5c\%<9c/ /g doesn't work
/\%c is a zero-width match.
You'll need to match something like:
/\v^(.).(.).(.)...
Which will keep the values of columns 1, 3, and 5 in groups.
Then you can substitute:
:%s!\v^(.).(.).(.)...!\1 \2 \3 !
...keeping columns 1, 3, and 5 but replacing the rest of the first eight columns with spaces.