CODEWITHSUNDEEP
python
James Hallen
James Hallen

Reputation: 5014

Creating list of dictionaries efficiently

I have implemented an approach to create a list of dictionaries, but I was wondering if there was a more efficient method of doing so (while preserving the order elements in both lists):

#global variable
dict_agriculture = []

col_name_agriculture = [tuple[0] for tuple in querycurs.description]
rows_agriculture = querycurs.fetchall()
for row in rows_agriculture:
    dict1 = dict(zip(col_name_agriculture, list(row)))
    dict_agriculture.append(dict1)

Upvotes: 1

Views: 137

Answers (3)

Aya
Aya

Reputation: 42090

I'm using sqlite.

In that case, the most efficient way would be to set row_factory to the built-in Row type, which is implemented in C, and will be faster than any equivalent code in Python.

For example...

import sqlite3

con = sqlite3.connect(...)
con.row_factory = sqlite3.Row
# Now queries will return objects which act as both a tuple, and a dictionary,
# so you can just do...
querycurs = con.cursor()
querycurs.execute(...)
dict_agriculture = querycurs.fetchall()

Upvotes: 2

Danny Staple
Danny Staple

Reputation: 7332

If I read this right, you have a query cursor, where the .description property is a list of tuples with field names as the first item. You then want to read the rows, turning each row list into a row dict using the row names from the tuple.

This can be done with a list comprehension:

col_name_agriculture = [tuple[0] for tuple in querycurs.description]
dict_agriculture = [dict(zip(col_name_agriculture, row) for row in querycurs]

Consider also how the final list will be used - a generator expression may be usable there if it is to be used in order.

Upvotes: 1

Martijn Pieters
Martijn Pieters

Reputation: 1124848

You could use a list comprehension here; its more efficient at creating the final list. Don't use .fetchall() when iteration will do, don't use tuple as a variable name, and there is no need to call list on each row:

fields = [tup[0] for tup in querycurs.description]
dict_agriculture = [dict(zip(fields, row)) for row in querycurs]

Upvotes: 4

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