Python ValueError when using SQLITE3 and dictionary

Question

After I was able to successfully store values from a SQLITE database in a python dictionary. However, when trying to process these values, I am getting the following error:

'Traceback (most recent call last):
   File "transportation problem.py", line 20, in <module>
   I = set([i for (i,k) in d])
ValueError: too many values to unpack

Does this indicate a problem with my source database, or is my code flawed?

import sqlite3 as lite
import sys
from gurobipy import *

con = lite.connect('transport.sqlite')

produce = {1:[2,4], 2:[1,2,3], 3:[2,3,4]}

d = {}
with con:

cur = con.cursor()    
cur.execute('SELECT idcustomer,idfactory,distance FROM distance')
result = cur.fetchall()
for idcustomer, idfactory, distance in result:
      d["({},{})".format(idcustomer, idfactory)] = distance
      I = set([i for (i,k) in d])
      K = set([k for (i,k) in d])
      J,M = multidict({1:3000, 2:3000, 3:3000})

Any help is highly appreciated!

Edit:　

The solution provided by Andy Hayden worked fine at first, but when I implemented it into my code, I started getting the following error:

c[i,j,k] = cost[i,j] * weight[k]
KeyError: ('1,1', 1)

The code I posted initially continues as follows:

 produce = {1:[2,4], 2:[1,2,3], 3:[2,3,4]}  
 K, weight = multidict({1:5, 2:2, 3:3, 4:4})            
      cost = {(1,1):4,  (1,2):6, (1,3):9,        
              (2,1):5,  (2,2):4, (2,3):7,
              (3,1):6,  (3,2):3, (3,3):4,
              (4,1):8,  (4,2):5, (4,3):3,
              (5,1):10, (5,2):8, (5,3):4,
              }
      c = {}
      for i in I:
        for j in J:
           for k in produce[j]:
               c[i,j,k] = cost[i,j] * weight[k]

Do I need to add '.iterititems' to the loop's items as well? When I input the SQLITE data manually, and revert 'd.iteritems' back to 'd', the code works without a problem.

Any suggestions would again be more than welcome!

Andy Hayden · Accepted Answer · 2014-01-23 04:40:25Z

1

d is a dictionary, so you need to iterate through its items for this to unpack correctly:

[i for (i, k) in d.items()]
[i for (i, k) in d.iteritems()]

By default you iterate over its keys:

for k in d:  # equivalent to for k in d.keys()

Though actually you don't need the list comprehension here:

I = set(d)  # equivalent to set(d.keys())

will give you the set of keys. Note: In python 3, d.keys() returns a set.

answered Jan 23, 2014 at 4:40

Andy Hayden

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