I have two text files Like
file1
1018 2
1019 7
1023 4
file2
1018 2
1019 7
1023 4
1026 8
I have a small bash code to find match and count
awk 'FNR==NR{a[$0]=1; next} $0 in a { count[$0]++ }
END { for( i in a ) print i, count[i]}' file1 file2
the output I get;
1018 2 1
1019 7 1
1023 4 1
I just want total count that is in this case: 3. Its simple to print count after the loop but didn't work, any solution....
When I have an outputted list in bash I'll use "wc". WC does a word count and you can specify it to count the number of lines. So say I want to count the number of files in a directory. I'll do:
ls -lh | wc -l
You could use a combination of sort and uniq to do so. This is how it would look like:
cat file1 file2 | sort | uniq -d | wc -l
Explanation:
cat is used to concatenate the two files
sort is used to sort the merged content
uniq (with option -d) is used to only show duplicated lines
wc (with option -l) is counting the left lines
Related
My text file would read as:
111
111
222
222
222
333
333
My resulting file would look like:
1,111
2,111
1,222
2,222
3,222
1,333
2,333
Or the resulting file could alternatively look like the following:
1
2
1
2
3
1
2
I've specified a comma as a delimiter here but it doesn't matter what the delimeter is --- I can modify that at a future date.In reality, I don't even need the original text file contents, just the line numbers, because I can just paste the line numbers against the original text file.
I am just not sure how I can go through numbering the lines based on repeated entries.
All items in list are duplicated at least once. There are no single occurrences of a line in the file.
$ awk -v OFS=',' '{print ++cnt[$0], $0}' file
1,111
2,111
1,222
2,222
3,222
1,333
2,333
Use a variable to save the previous line, and compare it to the current line. If they're the same, increment the counter, otherwise set it back to 1.
awk '{if ($0 == prev) counter++; else counter = 1; prev=$0; print counter}'
Perl solution:
perl -lne 'print ++$c{$_}' file
-n reads the input line by line
-l handles newlines
++$c{$_} increments the value assigned to the contents of the current line $_ in the hash table %c.
Software tools method, given textfile as input:
uniq -c textfile | cut -d' ' -f7 | xargs -L 1 seq 1
Shell loop-based variant of the above:
uniq -c textfile | while read a b ; do seq 1 $a ; done
Output (of either method):
1
2
1
2
3
1
2
I've a file (F1) with N=10000 lines, each line contains M=20000 numbers. I've an other file (F2) with N=10000 lines with only 1 column. How can count the number of occurences in line i of file F2 that are greater or equal to the number found at line i in the file F2 ? I tried using a bash loop with awk / sed but my output is empty.
Edit >
For now I've only succeed to print the number of occurences that are higher than a defined value. Here an example with a file with 3 lines and a defined value of 15 (sorry it's a very dirty code ..) :
for i in {1..3};do sed -n "$i"p tmp.txt | sed 's/\t/\n/g' | awk '{if($1 > 15){print $1}}' | wc -l; done;
Thanks in advance,
awk 'FNR==NR{a[FNR]=$1;next}
{count=0;for(i=1;i<=NF;i++)
{if($i >= a[FNR])
{count++}
};
print count
}' file2 file1
While processing file2, total line record is equal to line record of current file, store value in array a with current record as index.
initialize count to 0 for each line.
loop through the fields, increment the counter if value is greater or equal at current FNR index in array a.
Print the count value
$ cat file1
1 3 5 7 3 6
2 5 6 8 7 7
4 6 7 8 9 4
$ cat file2
6
3
1
$ awk -f file.awk
2
5
6
You could do it in a single awk command:
awk 'NR==FNR{a[FNR]=$1;next}{c=0;for(i=1;i<=NF;i++)c+=($i>a[FNR]);print c}' file2 file1
I have a big file with 1000 lines.I wanted to get 110 lines from it.
Lines should be evenly spread in Input file.
For example,I have read 4 lines from file with 10 lines
Input File
1
2
3
4
5
6
7
8
9
10
outFile:
1
4
7
10
Use:
sed -n '1~9p' < file
The -n option will stop sed from outputting anything. '1~9p' tells sed to print from line 1 every 9 lines (the p at the end orders sed to print).
To get closer to 110 lines you have to print every 9th line (1000/110 ~ 9).
Update: This answer will print 112 lines, if you need exactly 110 lines, you can limit the output just using head like this:
sed -n '1~9p' < file | head -n 110
$ cat tst.awk
NR==FNR { next }
FNR==1 { mod = int((NR-1)/tgt) }
!( (FNR-1)%mod ) { print; cnt++ }
cnt == tgt { exit }
$ wc -l file1
1000 file1
$ awk -v tgt=110 -f tst.awk file1 file1 > file2
$ wc -l file2
110 file2
$ head -5 file2
1
10
19
28
37
$ tail -5 file2
946
955
964
973
982
Note that this will not produce the output you posted in your question given your posted input file because that would require an algorithm that doesn't always use the same interval between output lines. You could dynamically calculate mod and adjust it as you parse your input file if you like but the above may be good enough.
With awk you can do:
awk -v interval=3 '(NR-1)%interval==0' file
where interval is the difference in line count between consecutive lines that are printed. The value is essentially a division of the total lines in the file divided by the number of lines that are printed.
I often like to use a combination of shell and awk for these sorts of things
#!/bin/bash
filename=$1
toprint=$2
awk -v tot=$(expr $(wc -l < $filename)) -v toprint=$toprint '
BEGIN{ interval=int((tot-1)/(toprint-1)) }
(NR-1)%interval==0 {
print;
nbr++
}
nbr==toprint{exit}
' $filename
Some examples:
$./spread.sh 1001lines 5
1
251
501
751
1001
$ ./spread.sh 1000lines 110 |head -n 3
1
10
19
$ ./spread.sh 1000lines 110 |tail -n 3
964
973
982
Say, I have two files and want to find out how many equal lines they have. For example, file1 is
1
3
2
4
5
0
10
and file2 contains
3
10
5
64
15
In this case the answer should be 3 (common lines are '3', '10' and '5').
This, of course, is done quite simply with python, for example, but I got curious about doing it from bash (with some standard utils or extra things like awk or whatever). This is what I came up with:
cat file1 file2 | sort | uniq -c | awk '{if ($1 > 1) {$1=""; print $0}}' | wc -l
It does seem too complicated for the task, so I'm wondering is there a simpler or more elegant way to achieve the same result.
P.S. Outputting the percentage of common part to the number of lines in each file would also be nice, though is not necessary.
UPD: Files do not have duplicate lines
To find lines in common with your 2 files, using awk :
awk 'a[$0]++' file1 file2
Will output 3 10 15
Now, just pipe this to wc to get the number of common lines :
awk 'a[$0]++' file1 file2 | wc -l
Will output 3.
Explanation:
Here, a works like a dictionary with default value of 0. When you write a[$0]++, you will add 1 to a[$0], but this instruction returns the previous value of a[$0] (see difference between a++ and ++a). So you will have 0 ( = false) the first time you encounter a certain string and 1 ( or more, still = true) the next times.
By default, awk 'condition' file is a syntax for outputting all the lines where condition is true.
Be also aware that the a[] array will expand every time you encounter a new key. At the end of your script, the size of the array will be the number of unique values you have throughout all your input files (in OP's example, it would be 9).
Note: this solution counts duplicates, i.e if you have:
file1 | file2
1 | 3
2 | 3
3 | 3
awk 'a[$0]++' file1 file2 will output 3 3 3 and awk 'a[$0]++' file1 file2 | wc -l will output 3
If this is a behaviour you don't want, you can use the following code to filter out duplicates :
awk '++a[$0] == 2' file1 file2 | wc -l
with your input example, this works too. but if the files are huge, I prefer the awk solutions by others:
grep -cFwf file2 file1
with your input files, the above line outputs
3
Here's one without awk that instead uses comm:
comm -12 <(sort file1.txt) <(sort file2.txt) | wc -l
comm compares two sorted files. The arguments 1,2 suppresses unique lines found in both files.
The output is the lines they have in common, on separate lines. wc -l counts the number of lines.
Output without wc -l:
10
3
5
And when counting (obviously):
3
You can also use comm command. Remember that you will have to first sort the files that you need to compare:
[gc#slave ~]$ sort a > sorted_1
[gc#slave ~]$ sort b > sorted_2
[gc#slave ~]$ comm -1 -2 sorted_1 sorted_2
10
3
5
From man pages for comm command:
comm - compare two sorted files line by line
Options:
-1 suppress column 1 (lines unique to FILE1)
-2 suppress column 2 (lines unique to FILE2)
-3 suppress column 3 (lines that appear in both files)
You can do all with awk:
awk '{ a[$0] += 1} END { c = 0; for ( i in a ) { if ( a[i] > 1 ) c++; } print c}' file1 file2
To get the percentage, something like this works:
awk '{ a[$0] += 1; if (NR == FNR) { b = FILENAME; n = NR} } END { c = 0; for ( i in a ) { if ( a[i] > 1 ) c++; } print b, c/n; print FILENAME, c/FNR;}' file1 file2
and outputs
file1 0.428571
file2 0.6
In your solution, you can get rid of one cat:
sort file1 file2| uniq -c | awk '{if ($1 > 1) {$1=""; print $0}}' | wc -l
How about keeping it nice and simple...
This is all that's needed:
cat file1 file2 | sort -n | uniq -d | wc -l
3
man sort:
-n, --numeric-sort -- compare according to string numerical value
man uniq:
-d, --repeated -- only print duplicate lines
man wc:
-l, --lines -- print the newline counts
Hope this helps.
EDIT - one fewer process (credit martin):
sort file1 file2 | uniq -d | wc -l
One way using awk:
awk 'NR==FNR{a[$0]; next}$0 in a{n++}END{print n}' file1 file2
Output:
3
The first answer by Aserre using awk is good but may have the undesirable effect of counting duplicates - even if the duplicates exist in only ONE of the files, which is not quite what the OP asked for.
I believe this edit will return only the unique lines that exist in BOTH files.
awk 'NR==FNR{a[$0]=1;next}a[$0]==1{a[$0]++;print $0}' file1 file2
If duplicates are desired, but only if they exist in both files, I believe this next version will work, but will only report duplicates in the second file that exist in the first file. (If the duplicates exist in the first file, only the those that also exist in file2 will be reported, so file order matters).
awk 'NR==FNR{a[$0]=1;next}a[$0]' file1 file2
Btw, I tried using grep, but it was painfully slow on files with a few thousand lines each. Awk is very fast!
UPDATE 1 : new version ensures intra-file duplicates are excluded from count, so only cross-file duplicates would show up in the final stats :
mawk '
BEGIN { _*= FS = "^$"
} FNR == NF { split("",___)
} ___[$_]++<NF { __[$_]++
} END { split("",___)
for (_ in __) {
___[__[_]]++ } printf(RS)
for (_ in ___) {
printf(" %\04715.f %s\n",_,___[_]) }
printf(RS) }' \
<( jot - 1 999 3 | mawk '1;1;1;1;1' | shuf ) \
<( jot - 2 1024 7 | mawk '1;1;1;1;1' | shuf ) \
<( jot - 7 1295 17 | mawk '1;1;1;1;1' | shuf )
3 3
2 67
1 413
===========================================
this is probably waaay overkill, but i wrote something similar to this to supplement uniq -c :
measuring the frequency of frequencies
it's like uniq -c | uniq -c without wasting time sorting. The summation and % parts are trivial from here, with 47 over-lapping lines in this example. It avoids spending any time performing per row processing, since the current setup only shows the summarized stats.
If you need to actual duplicated rows, they're also available right there serving as the hash key for the 1st array.
gcat <( jot - 1 999 3 ) <( jot - 2 1024 7 ) |
mawk '
BEGIN { _*= FS = "^$"
} { __[$_]++
} END { printf(RS)
for (_ in __) { ___[__[_]]++ }
for (_ in ___) {
printf(" %\04715.f %s\n",
_,___[_]) } printf(RS) }'
2 47
1 386
add another file, and the results reflect the changes (I added <( jot - 5 1295 5 ) ):
3 9
2 115
1 482
Given an input file containing one single number per line, how could I get a count of how many times an item occurred in that file?
cat input.txt
1
2
1
3
1
0
desired output (=>[1,3,1,1]):
cat output.txt
0 1
1 3
2 1
3 1
It would be great, if the solution could also be extended for floating numbers.
You mean you want a count of how many times an item appears in the input file? First sort it (using -n if the input is always numbers as in your example) then count the unique results.
sort -n input.txt | uniq -c
Another option:
awk '{n[$1]++} END {for (i in n) print i,n[i]}' input.txt | sort -n > output.txt
At least some of that can be done with
sort output.txt | uniq -c
But the order number count is reversed. This will fix that problem.
sort test.dat | uniq -c | awk '{print $2, $1}'
Using maphimbu from the Debian stda package:
# use 'jot' to generate 100 random numbers between 1 and 5
# and 'maphimbu' to print sorted "histogram":
jot -r 100 1 5 | maphimbu -s 1
Output:
1 20
2 21
3 20
4 21
5 18
maphimbu also works with floating point:
jot -r 100.0 10 15 | numprocess /%10/ | maphimbu -s 1
Output:
1 21
1.1 17
1.2 14
1.3 18
1.4 11
1.5 19
In addition to the other answers, you can use awk to make a simple graph. (But, again, it's not a histogram.)
perl -lne '$h{$_}++; END{for $n (sort keys %h) {print "$n\t$h{$n}"}}' input.txt
Loop over each line with -n
Each $_ number increments hash %h
Once the END of input.txt has been reached,
sort {$a <=> $b} the hash numerically
Print the number $n and the frequency $h{$n}
Similar code which works on floating point:
perl -lne '$h{int($_)}++; END{for $n (sort {$a <=> $b} keys %h) {print "$n\t$h{$n}"}}' float.txt
float.txt
1.732
2.236
1.442
3.162
1.260
0.707
output:
0 1
1 3
2 1
3 1
I had a similar problem as described, but across gigabytes of gzip'd log files. Because many of these solutions necessitated waiting until all the data was parsed, I opted to write rare to quickly parse and aggregate data based on a regexp.
In the case above, it's as simple as passing in the data to the histogram function:
rare histo input.txt
# OR
cat input.txt | rare histo
# Outputs:
1 3
0 1
2 1
3 1
But it can also handle more complex cases via regex/expressions, such as:
rare histo --match "(\d+)" --extract "{1}" input.txt