I have a text file : ABC.txt which contain below data
A Apple a day keeps a doctor away
B I like to play with Ball
C I have cat at my home
D My Dog name is bob
I want to display output on my screen with 10 spaces in a frontend and then my file data
Expected output :
A Apple a day keeps a doctor away
B I like to play with Ball
C I have cat at my home
D My Dog name is bob
I have tried this but not working
Command :
cat ABC.txt | column -t
prefix=' '
sed "s/^/$prefix/" ABC.txt
^ matches the beginning of the line, and this replaces it with 10 spaces.
If the number of spaces can vary, you can calculate the prefix variable instead of hard-coding it. See How can I repeat a character in Bash?
Related
My text file should be of two columns separated by a tab-space (represented by \t) as shown below. However, there are a few corrupted values where column 1 has two values separated by a space (represented by \s).
A\t1
B\t2
C\sx\t3
D\t4
E\sy\t5
My objective is to create a table as follows:
A\t1
B\t2
C\t3
D\t4
E\t5
i.e. discard the 2nd value that is present after the space in column 1 for eg. in C\sx\t3 I can discard the x that is present after space and store the columns as C\t3.
I have tried a couple of things but with no luck.
I tried to cut the cols based on \t into independent columns and then cut the first column based on \s and join them again. However, it did not work.
Here is the snippet:
col1=(cut -d$'\t' -f1 $file | cut -d' ' -f1)
col2=(cut -d$'\t' -f1 $file)
myArr=()
for((idx=0;idx<${#col1[#]};idx++));do
echo "#{col1[$idx]} #{col2[$idx]}"
# I will append to myArr here
done
The output is appending the list of col2 to the col1 as A B C D E 1 2 3 4 5. And on top of this, my file is very huge i.e. 5,300,000 rows so I would like to avoid looping over all the records and appending them one by one.
Any advice is very much appreciated.
Thank you. :)
And another sed solution:
Search and replace any literal space followed by any number of non-TAB-characters with nothing.
sed -E 's/ [^\t]+//' file
A 1
B 2
C 3
D 4
E 5
If there could be more than one actual space in there just make it 's/ +[^\t]+//' ...
Assuming that when you say a space you mean a blank character then using any awk:
awk 'BEGIN{FS=OFS="\t"} {sub(/ .*/,"",$1)} 1' file
Solution using Perl regular expressions (for me they are easier than seds, and more portable as there are few versions of sed)
$ cat ls
A 1
B 2
C x 3
D 4
E y 5
$ cat ls |perl -pe 's/^(\S+).*\t(\S+)/$1 $2/g'
A 1
B 2
C 3
D 4
E 5
This code gets all non-empty characters from the front and all non-empty characters from after \t
Try
sed $'s/^\\([^ \t]*\\) [^\t]*/\\1/' file
The ANSI-C Quoting ($'...') feature of Bash is used to make tab characters visible as \t.
take advantage of FS and OFS and let them do all the hard work for you
{m,g}awk NF=NF FS='[ \t].*[ \t]' OFS='\t'
A 1
B 2
C 3
D 4
E 5
if there's a chance of leading edge or trailing edge spaces and tabs, then perhaps
mawk 'NF=gsub("^[ \t]+|[ \t]+$",_)^_+!_' OFS='\t' RS='[\r]?\n'
Hi I'm trying to solve a problem only using sed commands and without using pipeline. But I am allowed to pass the result of a sed command to a file or te read from a file.
EX:
sed s/dog/cat/ >| tmp
or
sed s/dog/cat/ < tmp
Anyway lets say I had a file F1 and its contents was :
Hello hi 123
if a equals b
you
one abc two three four
dany uri four 123
The output should be:
if if if a equals b
dany dany dany uri four 123
Explanation: the program must only print lines that have exactly 4 words and when it prints them it must print the first word of the line 3 times.
I've tried doing commands like this:
sed '/[^ ]*.[^ ]*.[^ ]*/s/[^ ]\+/& & &/' F1
or
sed 's/[^ ]\+/& & &/' F1
but I can't figure out how i can calculate with sed that there are only 4 words in a line.
any help will be appreciated
$ sed -En 's/^([^[:space:]]+)([[:space:]]+[^[:space:]]+){3}$/\1 \1 &/p' file
if if if a equals b
dany dany dany uri four 123
The above uses a sed that supports EREs with a -E option, e.g. GNU and OSX seds).
If the fields are tab separated
sed 'h;s/[^[:blank:]]//g;s/[[:blank:]]\{3\}//;/^$/!d;x;s/\([^[:blank:]]*[[:blank:]]\)/\1\1\1/' infile
I want to remove certain lines from a tab-delimited file and write output to a new file.
a b c 2017-09-20
a b c 2017-09-19
es fda d 2017-09-20
es fda d 2017-09-19
The 4th column is Date, basically I want to keep only lines that has 4th column as "2017-09-19" (keep line 2&4) and write to a new file. The new file should have same format as the raw file.
How to write the linux command for this example?
Note: The search criteria should be on the 4th field as I have other fields in the real data and possibly have same value as 4th field.
With awk:
awk 'BEGIN{OFS="\t"} $4=="2017-09-19"' file
OFS: output field separator, a space by default
Use grep to filter:
cat file.txt | grep '2017-09-19' > filtered_file.txt
This is not perfect, since the string 2017-09-19 is not required to appear in the 4th column, but if your file looks like the example, it'll work.
Sed solution:
sed -nr "/^([^\t]*\t){3}2017-09-19/p" input.txt >output.txt
this is:
-n - don't output every line
-r - extended regular expresion
/regexp/p - print line that contains regular expression regexp
^ - begin of line
(regexp){3} - repeat regexp 3 times
[^\t] - any character except tab
\t - tab character
* - repeat characters multiple times
2017-09-19 - search text
That is, skip 3 columns separated by a tab from the beginning of the line, and then check that the value of column 4 coincides with the required value.
awk '/2017-09-19/' file >newfile
cat newfile
a b c 2017-09-19
es fda d 2017-09-19
Can someone help me how to write a piece of command that will insert some text in multiple places (given column and row) of a given file that already contains data. For example: old_data is a file that contains:
A
And I wish to get new_data that will contain:
A 1
I read something about awk and sed commands, but I don't believe to understand how to incorporate these, to get what I want.
I would like to add up, that this command I would like to use as a part of script
for b in ./*/ ; do (cd "$b" && command); done
If we imagine content of old_data as a matrix of elements {An*m} where n corresponds to number of row and m to number of column of this matrix, I wish to manipulate with matrix so that I could add new elements. A in old-data has coordinates (1,1). In new_data therefore, I wish to assign 1 to a matrix element that has coordinates (1,3).
If we compare content of old_data and new_data we see that (1,2) element corresponds to space (it is empty).
It's not at all clear to me what you are asking for, but I suspect you are saying that you would like a way to insert some given text in to a particular row and column. Perhaps:
$ cat input
A
B
C
D
$ row=2 column=2 text="This is some new data"
$ awk 'NR==row {$column = new_data " " $column}1' row=$row column=$column new_data="$text" input
A
B This is some new data
C
D
This bash & unix tools code works:
# make the input files.
echo {A..D} | tr ' ' '\n' > abc ; echo {1..4} | tr ' ' '\n' > 123
# print as per previous OP spec
head -1q abc 123 ; paste abc 123 123 | tail -n +2
Output:
A
1
B 2 2
C 3 3
D 4 4
Version #3, (using commas as more visible separators), as per newest OP spec:
# for the `sed` code change the `2` to whatever column needs deleting.
paste -d, abc 123 123 | sed 's/[^,]*//2'
Output:
A,,1
B,,2
C,,3
D,,4
The same, with tab delimiters (less visually obvious):
paste abc 123 123 | sed 's/[^\t]*//2'
A 1
B 2
C 3
D 4
I'm using grep to display all lines that have ONLY 4,5,6,7 and 9 in the zipcode column.
How do i display only the lines of the file that contain the numbers 4,5,6,7 and 9 in the zipcode field?
A sample row is:
15 m jagger mick 41 4th 95115
Thanks
I am going to assume you meant "How do I use grep to..."
If all of the lines in the file have a 5 digit zip at the end of each line, then:
egrep "[45679]{5}$" filename
Should give you what you want.
If there might be whitespace between the zip and the end of the line, then:
egrep "[45679]{5}[[:space:]]*$" filename
would be more robust.
If the problem is more general than that, please describe it more accurately.
Following regex should fetch you desired result:
egrep "[45679]+$" file
If by "grep" you mean, "the correct tool", then the solution you seek is:
awk '$7 ~ /^[45679]*$/' input
This will print all lines of input in which the 7th field consists only of the characters 4,5,6,7, and 9. If you want to specify 'the last column' rather than the 7th, try
awk '$NF ~ /^[45679]*$/' input