create a dict by spliting a string in an ordered dict

Question

I have an ordered dict that represent field definition ie Name, type, width, precision

it looks like this:

 OrderedDict([(u'CODE_MUN', 'str:8'), (u'CODE_DR_AN', 'str:8'),
 (u'AREA', 'float:31.2'), (u'PERIMETER', 'float:31.4')])

I would like to create a dict for each item that would be like this:

{'name' : 'CODE_MUN', 'type': 'str', 'width': 8, 'precision':0} for fields without precision

and

{'name' : 'AREA', 'type': 'float', 'width': 31, 'precision':2 } for fiels with precision

for keys, values in fieldsDict.iteritems():
   dict = {}
   dict['name'] = keys
   props = re.split(':.', values)
   dict['type'] = props[0]
   dict['width'] = props[1]
   dict['precision'] = props[2]

of course I have index error when there is no precision defined. What would be the best way to achieve that?

Answer 1

You have to check precision is there or not.

from collections import OrderedDict
import re

fieldsDict = OrderedDict([(u'CODE_MUN', 'str:8'), (u'CODE_DR_AN', 'str:8'),
 (u'AREA', 'float:31.2'), (u'PERIMETER', 'float:31.4')])

for keys, values in fieldsDict.iteritems():
   dict = {}
   dict['name'] = keys
   props = re.split(':.', values)
   dict['type'] = props[0]
   dict['width'] = props[1]
   if len(props) == 3:
       dict['precision'] = props[2]
   else:
       dict['precision'] = 0
   print dict

This might be help

Answer 2

Use a try-except block.

for keys, values in fieldsDict.iteritems():
    dict = {}
    dict['name'] = keys
    props = re.split(':.', values)
    dict['type'] = props[0]
    dict['width'] = props[1]

    try:
        dict['precision'] = props[2]
    except IndexError:
        dict['precision'] = 0

You could also test for length using an if-else block. The methods are pretty close and I doubt this is a situation where it really matters, but for more on asking forgiveness vs permission you can see this question .