Question

Afternoon, i am trying to retrieve seat numbers from a database using the following code

        cur.execute("SELECT * FROM seats")
    while True:
        row = cur.fetchone()
        if row == None:
            break
        print row[0]

But when i do so, it prints out each individual record one per line like so :

A1
A2
B3 etc..

But i want each row to print out with the same letter if that makes sense such as :

A1 A2 A3 A4 A5 A6 A7 A8
B1 B2 B3 B4 B5 B6 B7

But i cant seem to get it like that ? How would i go about doing this?

Answer 1

Use the itertools.groupby() tool:

from itertools import groupby

for letter, rows in groupby(cur, key=lambda r: r[0][0]):
    print ' '.join([r[0] for r in rows])

The groupby() function loops over each row in cur, take the first letter of the first column, and give you a tuples with each (letter, rows) values. The rows value is another iterable, you can loop over that (with a for loop, for example) to list all rows that have that first letter.

This does rely on the rows being sorted already. If your rows alternate between first letters:

A1
A2
B1
B2
A3
A4

it’ll print those as separate groups:

A1 A2
B1 B2
A3 A4

You may want to add a ORDER BY firstcolumnname ordering instruction to your query to ensure correct grouping.

This is what I see when I create a test db:

>>> cur.execute("SELECT * FROM seats ORDER BY code")
<sqlite3.Cursor object at 0x10b1a8730>
>>> for letter, rows in groupby(cur, key=lambda r: r[0][0]):
...     print ' '.join([r[0] for r in rows])
... 
A1 A2 A3 A4 A5 A6 A7 A8
B1 B2 B3 B4 B5 B6 B7 B8
C1 C2 C3 C4 C5 C6 C7 C8