How to use "column" to center a chart? - bash

Question

I was wondering what the best way to sort a chart using the column command to center each column instead of the default left aligned column was. I have been using the column -t filename command.
Current Output:
Label1 label2
Anotherlabel label2442
label152 label42242
label78765 label373737737
Desired Output: Something like this
Label1 label2
Anotherlabel label2442
label152 label42242
label78765 label373737737
Basically, I want it to be centered instead of left aligned.

Answer 1

Here is an awk solution:
# Collect all lines in "data", keep track of maximum width for each field
{
data[NR] = $0
for (i = 1; i <= NF; ++i)
max[i] = length($i) > max[i] ? length($i) : max[i]
}
END {
for (i = 1; i <= NR; ++i) {
# Split record into array "arr"
split(data[i], arr)
# Loop over array
for (j = 1; j <= NF; ++j) {
# Calculate amount of padding required
pad = max[j] - length(arr[j])
# Print field with appropriate padding, see below
printf "%*s%*s%s", length(arr[j]) + int(pad/2), arr[j], \
pad % 2 == 0 ? pad/2 : int(pad/2) + 1, "", \
j == NF ? "" : " "
}
# Newline at end of record
print ""
}
}
Called like this:
$ awk -f centre.awk infile
Label1 label2
Anotherlabel label2442
label152 label42242
label78765 label373737737
The printf statement uses padding with dynamic widths:
The first %*s takes care of left padding and the data itself: arr[j] gets printed and padded to a total width of length(arr[j]) + int(pad/2).
The second %*s prints the empty string, left padded to half of the total padding required. pad % 2 == 0 ? pad/2 : int(pad/2) + 1 checks if the total padding was an even number, and if not, adds an extra space.
The last %s prints j == NF ? "" : " ", i.e., two spaces, unless we're at the last field.
Some older awks don't support the %*s syntax, but the formatting string can be assembled like width = 5; "%" width "s" in that case.

Answer 2

Here's a Python program to do what you want. It's probably too hard to do in bash, so you'll need to use a custom program or awk script. Basic algorithm:
count number of columns
[optional] make sure each line has the same number of columns
figure out the maximum length of data for each column
print each line using the max lengths
.
#!/usr/bin/env python3
import sys
def column():
# Read file and split each line into fields (by whitespace)
with open(sys.argv[1]) as f:
lines = [line.split() for line in f]
# Check that each line has the same number of fields
num_fields = len(lines[0])
for n, line in enumerate(lines):
if len(line) != num_fields:
print('Line {} has wrong number of columns: expected {}, got {}'.format(n, num_fields, len(line)))
sys.exit(1)
# Calculate the maximum length of each field
max_column_widths = [0] * num_fields
for line in lines:
line_widths = (len(field) for field in line)
max_column_widths = [max(z) for z in zip(max_column_widths, line_widths)]
# Now print them centered using the max_column_widths
spacing = 4
format_spec = (' ' * spacing).join('{:^' + str(n) + '}' for n in max_column_widths)
for line in lines:
print(format_spec.format(*line))
if __name__ == '__main__':
column()