mjhjmtu

I have the following output:

2015/1/7 8
2015/1/8 49
2015/1/9 40
2015/1/10 337
2015/1/11 11
2015/1/12 3
2015/1/13 9
2015/1/14 102
2015/1/15 62
2015/1/16 10
2015/1/17 30
2015/1/18 30
2015/1/19 1
2015/1/20 3
2015/1/21 23
2015/1/22 12
2015/1/24 6
2015/1/25 3
2015/1/27 2
2015/1/28 16
2015/1/29 1
2015/2/1 12
2015/2/2 2
2015/2/3 1
2015/2/4 10
2015/2/5 13
2015/2/6 2
2015/2/9 2
2015/2/10 25
2015/2/11 1
2015/2/12 6
2015/2/13 12
2015/2/14 2
2015/2/16 8
2015/2/17 8
2015/2/20 1
2015/2/23 1
2015/2/27 1
2015/3/2 3
2015/3/3 2

And I'd like to draw a histogram

2015/1/7 ===
2015/1/8 ===========
2015/1/9 ==========
2015/1/10 ====================================================================
2015/1/11 ===
2015/1/11 =
...

Do you know if there is a bash command that would let me do that?

Try this in perl :

perl -lane 'print $F[0], "t", "=" x ($F[1] / 5)' file

EXPLANATIONS:

• 
  -a is an explicit split() in @F array, we get the values with $F[n]
  


• 
  x is to tell perl to print a character N times


• 
  ($F[1] / 5) : here we get the number and divide it by 5 for a pretty print output

In perl:

perl -pe 's/ (d+)$/"="x$1/e' file

• 
  e causes the expression to be evaluated, so I get = repeated using the value of $1 (the number matched by (d+)).


• You could do "="x($1/3) instead of "="x$1 to get shorter lines. (The / is escaped since we're in the middle of a substitution command.)

In bash (inspired from this SO answer):

while read d n 
do 
 printf "%st%$nsn" "$d" = | tr ' ' '=' 
done < test.txt

• 
  printf pads the second string using spaces to get a width of $n (%$ns), and I replace the spaces with =.


• The columns are delimited using a tab (t), but you can make it prettier by piping to column -ts't'.


• You could use $((n/3)) instead of $n to get shorter lines.

Another version:

unset IFS; printf "%st%*sn" $(sed 's/$/ =/' test.txt) | tr ' ' =

The only drawback I can see is that you'll need to pipe sed's output to something if you want to scale down, otherwise this is the cleanest option. If there is a chance of your input file containing one of [?* you should lead the command w/ set -f;.

Easy with awk

awk '$2=sprintf("%-*s", $2, ""); gsub(" ", "=", $2); printf("%-10s%sn", $1, $2)' file

2015/1/7 ========
2015/1/8 =================================================
2015/1/9 ========================================
..
..

Or with my favourite programming language

python3 -c 'import sys
for line in sys.stdin:
 data, width = line.split()
 print(":<10:=<width".format(data, "", width=width))' <file

How about:

#! /bin/bash
histo="======================================================================+"

read datewd value

while [ -n "$datewd" ] ; do
 # Use a default width of 70 for the histogram
 echo -n "$datewd "
 echo $histo:0:$value

 read datewd value
done

Which produces:

~/bash $./histogram.sh < histdata.txt
2015/1/7 ========
2015/1/8 =================================================
2015/1/9 ========================================
2015/1/10 ======================================================================+
2015/1/11 ===========
2015/1/12 ===
2015/1/13 =========
2015/1/14 ======================================================================+
2015/1/15 ==============================================================
2015/1/16 ==========
2015/1/17 ==============================
2015/1/18 ==============================
2015/1/19 =
2015/1/20 ===
2015/1/21 =======================
2015/1/22 ============
2015/1/24 ======
2015/1/25 ===
2015/1/27 ==
2015/1/28 ================
2015/1/29 =
2015/2/1 ============
2015/2/2 ==
2015/2/3 =
2015/2/4 ==========
2015/2/5 =============
2015/2/6 ==
2015/2/9 ==
2015/2/10 =========================
2015/2/11 =
2015/2/12 ======
2015/2/13 ============
2015/2/14 ==
2015/2/16 ========
2015/2/17 ========
2015/2/20 =
2015/2/23 =
2015/2/27 =
2015/3/2 ===
2015/3/3 ==
~/bash $

This struck me as a fun traditional command line problem. Here's my bash script solution:

awk 'if (count[$1])count[$1] += $2 else count[$1] = $2 
 ENDfor (year in count) print year, count[year];' data |
sed -e 's/// /g' | sort -k1,1n -k2,2n -k3,3n |
awk 'printf("%d/%d/%dt", $1,$2,$3); for (i=0;i<$4;++i) printf("="); printf("n");'

The little script above assumes the data is in a file imaginatively named "data".

I'm not too happy with the "run it through sed and sort" line - it would be unnecessary if your month and day-of-month always had 2 digits, but that's life.

Also, as a historical note, traditional Unixes used to come with a command line plotting utility that could do fairly ugly ASCII graphs and plots. I can't remember the name, but it looks like GNU plotutils replace the old traditional utility.