从字段：值格式转换为CSV

Question

I have a file in the following format (well, sort of):

RECORD_SEPARATOR
foo: some foo value
bar: another value
baz: 123
RECORD_SEPARATOR
foo: another foo value
bar: yet another value
baz: 345
RECORD_SEPARATOR
foo: a third foo
RECORD_SEPARATOR
bar: a fourth bar
baz: 111

and so on. The key point here is that not all records have all fields present.

My question: What's a super-simple way to convert this data into CSV format? That is, in my example

foo,bar,baz
some foo value,another value,123
another foo value,yet another value,345
a third foo,,
,a fourth bar,111

Of course you can write a awk (or perl, or Python) script for this, but I'm hoping there's something pre-existing, or some trick to make it a very short script.

Note: I'm looking for something that's Unix-command-line-oriented of course.

Answer 1

Hi with the great Miller http://johnkerl.org/miller/doc , starting from

foo: some foo value
bar: another value
baz: 123

foo: another foo value
bar: yet another value
baz: 345

foo: a third foo

bar: a fourth bar
baz: 111

you can run

mlr --x2p --ips ": " --barred cat then unsparsify --fill-with "" inputFile

and have this pretty print output

+-------------------+-------------------+-----+
| foo               | bar               | baz |
+-------------------+-------------------+-----+
| some foo value    | another value     | 123 |
| another foo value | yet another value | 345 |
| a third foo       | -                 | -   |
| -                 | a fourth bar      | 111 |
+-------------------+-------------------+-----+

If you want a CSV, run

mlr --x2c --ips ": " cat then unsparsify --fill-with "" inputFile

and you will have

foo,bar,baz
some foo value,another value,123
another foo value,yet another value,345
a third foo,,
,a fourth bar,111