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0

Given the following list (from an sqlite query):

[('job1', 'location1', 10), 
('job1', 'location2', 10),
('job2', 'location1', 5),
('job3', 'location1', 10),
('job3', 'location3', 10)]

I'd like to have the following rendered in my tpl template:

job1
location1: 10
location2: 10

job2
location1: 5

job3
location1: 10
location3: 10

I can probably get this done with setdefault

d = 

for job in jobs:
 d.setdefault(job[0], ).update(job[1]: job[2])

but I'm wondering what is the standard or best practice way of doing this?

Cheers,

python sqlite bottle

share|improve this question

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

add a comment | 

0

Given the following list (from an sqlite query):

[('job1', 'location1', 10), 
('job1', 'location2', 10),
('job2', 'location1', 5),
('job3', 'location1', 10),
('job3', 'location3', 10)]

I'd like to have the following rendered in my tpl template:

job1
location1: 10
location2: 10

job2
location1: 5

job3
location1: 10
location3: 10

I can probably get this done with setdefault

d = 

for job in jobs:
 d.setdefault(job[0], ).update(job[1]: job[2])

but I'm wondering what is the standard or best practice way of doing this?

Cheers,

python sqlite bottle

share|improve this question

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

add a comment | 

Given the following list (from an sqlite query):

[('job1', 'location1', 10), 
('job1', 'location2', 10),
('job2', 'location1', 5),
('job3', 'location1', 10),
('job3', 'location3', 10)]

I'd like to have the following rendered in my tpl template:

job1
location1: 10
location2: 10

job2
location1: 5

job3
location1: 10
location3: 10

I can probably get this done with setdefault

d = 

for job in jobs:
 d.setdefault(job[0], ).update(job[1]: job[2])

but I'm wondering what is the standard or best practice way of doing this?

Cheers,

python sqlite bottle

share|improve this question

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

[('job1', 'location1', 10), 
('job1', 'location2', 10),
('job2', 'location1', 5),
('job3', 'location1', 10),
('job3', 'location3', 10)]

job1
location1: 10
location2: 10

job2
location1: 5

job3
location1: 10
location3: 10

d = 

for job in jobs:
 d.setdefault(job[0], ).update(job[1]: job[2])

share|improve this question

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

share|improve this question

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

asked Mar 23 at 19:34

S1M0N_HS1M0N_H

3225

1 Answer
1

active

oldest

votes

1

Here's how I'd make your code more Pythonic:

from collections import defaultdict

d = defaultdict(dict)

for (job_id, location, value) in jobs:
 d[job_id][location] = value

# if you need an actual dict at the end (and not a defaultdict),
# use d = dict(d)

What I changed:

1. Use a defaultdict.


2. Use tuple unpacking for additional readability.

share|improve this answer

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

add a comment | 

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1

Here's how I'd make your code more Pythonic:

from collections import defaultdict

d = defaultdict(dict)

for (job_id, location, value) in jobs:
 d[job_id][location] = value

# if you need an actual dict at the end (and not a defaultdict),
# use d = dict(d)

What I changed:

1. Use a defaultdict.


2. Use tuple unpacking for additional readability.

share|improve this answer

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

add a comment | 

1

Here's how I'd make your code more Pythonic:

from collections import defaultdict

d = defaultdict(dict)

for (job_id, location, value) in jobs:
 d[job_id][location] = value

# if you need an actual dict at the end (and not a defaultdict),
# use d = dict(d)

What I changed:

1. Use a defaultdict.


2. Use tuple unpacking for additional readability.

share|improve this answer

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

add a comment | 

Here's how I'd make your code more Pythonic:

from collections import defaultdict

d = defaultdict(dict)

for (job_id, location, value) in jobs:
 d[job_id][location] = value

# if you need an actual dict at the end (and not a defaultdict),
# use d = dict(d)

What I changed:

1. Use a defaultdict.


2. Use tuple unpacking for additional readability.

share|improve this answer

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

from collections import defaultdict

d = defaultdict(dict)

for (job_id, location, value) in jobs:
 d[job_id][location] = value

# if you need an actual dict at the end (and not a defaultdict),
# use d = dict(d)

share|improve this answer

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831

answered Mar 25 at 13:19

ron rothmanron rothman

10.8k42831