I have a column which has "--" pattern in the beginning, middle and end of the string. For example:
-- myString
my -- String
myString --
I want to find these two types of cells
-- myString
myString --
and remove the "--" pattern, so it will look fine! I am an amateur user of excel but can use functions if you suggest me. It should be possible with find and use the results of the Find in Replace functions, but I do not know how to pass the results to Replace.
Please note: The answer should take care all the cells in the column, which are hundreds. One solution for changing all, not one solution for one cell.
EDIT: Just reread the request, per instruction from Gary'sStudent. This will remove all instances of "--", not only those at the beginning/end.
If the data is in A1, use the following formula:
=SUBSTITUTE(A1,"--","")
With data in A1 in B1 enter:
=IF(LEFT(A1,2)="--",MID(A1,3,9999),IF(RIGHT(A1,2)="--",MID(A1,1,LEN(A1)-2),A1))
OK, I found the answer. The answer from #Dubison helped me to find the right answer.
If the left two characters in this cell is "--" and the last two characters are "--" the substitute the "--" with "", else to nothing.
=IF(LEFT(A1,2)="--",SUBSTITUTE(A1,"--",""),IF(RIGHT(A1,2)="--",SUBSTITUTE(A1,"--",""), A1))
This will be pretty much the same with previous answers, only using simpler logic. If your strings first or last character = "-" do nothing, else replace "--" with "".
=IF(LEFT(A1,1)="-",A1,IF(RIGHT(A1,1)="-",A1, SUBSTITUTE(A1,"--","")))
UPDATE:
I noticed that I have misread the question. Above code will remove the "--" only if it is in the middle. However original question was to remove "--" only if it is at the beginning or at the end. So formula should be:
=IF(OR(LEFT(A1,2)="--",RIGHT(A1,2)="--"),SUBSTITUTE(A1,"--",""),A1)
I want to add a minus sign "-" infront of the first word in a line on the editor VIM. The lines contains spaces for indentation. The indentation shall not be touched. E.g
As Is
list point 1
sub list point 2
and so on...
I want
- list point 1
- sub list point 2
- and so on...
I can find the first word, but i struggle with replacing it in the correct way.
^\s*\w
in Vim
/^\s*\w
But in the replacement I always remove the complete found part....
:s/^\s*\w/- \w/
Which leads to
- ist point 1
- ub list point 2
- nd so on...
Use & which is replaced with the matched string:
:%s/\w/- &
I'm late to the party but:
:%norm! I- <CR>
And another one with :s:
:%s/^\s*/&- /
An alternative to falsetrue's answer: You can capture the first word character and print it out along with the leading -:
%s/\(\w\)/- \1/
:normal cmd may help too:
:%norm! wi-
note that after - there is a space.
I have a data file as follows.
1,14.23,1.71,2.43,15.6,127,2.8,3.06,.28,2.29,5.64,1.04,3.92,1065
1,13.2,1.78,2.14,11.2,100,2.65,2.76,.26,1.28,4.38,1.05,3.4,1050
1,13.16,2.36,2.67,18.6,101,2.8,3.24,.3,2.81,5.68,1.03,3.17,1185
1,14.37,1.95,2.5,16.8,113,3.85,3.49,.24,2.18,7.8,.86,3.45,1480
1,13.24,2.59,2.87,21,118,2.8,2.69,.39,1.82,4.32,1.04,2.93,735
Using vim, I want to reomve the 1's from each of the lines and append them to the end. The resultant file would look like this:
14.23,1.71,2.43,15.6,127,2.8,3.06,.28,2.29,5.64,1.04,3.92,1065,1
13.2,1.78,2.14,11.2,100,2.65,2.76,.26,1.28,4.38,1.05,3.4,1050,1
13.16,2.36,2.67,18.6,101,2.8,3.24,.3,2.81,5.68,1.03,3.17,1185,1
14.37,1.95,2.5,16.8,113,3.85,3.49,.24,2.18,7.8,.86,3.45,1480,1
13.24,2.59,2.87,21,118,2.8,2.69,.39,1.82,4.32,1.04,2.93,735,1
I was looking for an elegant way to do this.
Actually I tried it like
:%s/$/,/g
And then
:%s/$/^./g
But I could not make it to work.
EDIT : Well, actually I made one mistake in my question. In the data-file, the first character is not always 1, they are mixture of 1, 2 and 3. So, from all the answers from this questions, I came up with the solution --
:%s/^\([1-3]\),\(.*\)/\2,\1/g
and it is working now.
A regular expression that doesn't care which number, its digits, or separator you've used. That is, this would work for lines that have both 1 as their first number, or 114:
:%s/\([0-9]*\)\(.\)\(.*\)/\3\2\1/
Explanation:
:%s// - Substitute every line (%)
\(<something>\) - Extract and store to \n
[0-9]* - A number 0 or more times
. - Every char, in this case,
.* - Every char 0 or more times
\3\2\1 - Replace what is captured with \(\)
So: Cut up 1 , <the rest> to \1, \2 and \3 respectively, and reorder them.
This
:%s/^1,//
:%s/$/,1/
could be somewhat simpler to understand.
:%s/^1,\(.*\)/\1,1/
This will do the replacement on each line in the file. The \1 replaces everything captured by the (.*)
:%s/1,\(.*$\)/\1,1/gc
.........................
You could also solve this one using a macro. First, think about how to delete the 1, from the start of a line and append it to the end:
0 go the the start of the line
df, delete everything to and including the first ,
A,<ESC> append a comma to the end of the line
p paste the thing you deleted with df,
x delete the trailing comma
So, to sum it up, the following will convert a single line:
0df,A,<ESC>px
Now if you'd like to apply this set of modifications to all the lines, you will first need to record them:
qj start recording into the 'j' register
0df,A,<ESC>px convert a single line
j go to the next line
q stop recording
Finally, you can execute the macro anytime you want using #j, or convert your entire file with 99#j (using a higher number than 99 if you have more than 99 lines).
Here's the complete version:
qj0df,A,<ESC>pxjq99#j
This one might be easier to understand than the other solutions if you're not used to regular expressions!