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pythonordereddictionary

OrderedDict Merge and Append 0 if not

from collections import OrderedDict
b= OrderedDict([('04-25', 5), ('04-23', 53), ('04-20', 3), ('04-2', 33)])
# b = OrderedDict()
c = OrderedDict([('04-20', 5), ('03-18', 53), ('03-13', 3), ('03-12', 133)])
b_data = OrderedDict()
for k , v in b.items()+c.items():
    b_data.setdefault(k,[]).append(v)

f_data = OrderedDict()

for k1, v1 in b_data.iteritems():
    if len(v1)==1:
        f_data[k1]=v1[0],0
    else:
        f_data[k1]=v1[0],v1[1]
print f_data

I can get the output of f_data 

OrderedDict([('04-25', (5, 0)), ('04-23', (53, 0)), ('04-20', (3, 5)), ('04-2', (33, 0)), ('03-18', (53, 0)), ('03-13', (3, 0)), ('03-12', (133, 0))])

But while merging b and c items , In f_data value[0] should be b value and value[1] should be c value. if there is no value put 0 in that.

If the b items is empty also we have to keep 0 in value[0] and c items in value[1].

If b and c have same key merge along with their values as I mentioned above.

My expected output for the above data is ,

OrderedDict([('04-25', (5, 0)), ('04-23', (53, 0)), ('04-20', (3, 5)), ('04-2', (33, 0)), ('03-18', (0, 53)), ('03-13', (0, 3)), ('03-12', (0,133))])
Solution

  • This will do the job for you:

    from collections import OrderedDict

b = OrderedDict([('04-25', 5), ('04-23', 53), ('04-20', 3), ('04-2', 33)])
c = OrderedDict([('04-20', 5), ('03-18', 53), ('03-13', 3), ('03-12', 133)])    
f_data = OrderedDict()
for key in b.keys() + c.keys():
    if not f_data.has_key(key):
        try:
            val_b = b[key]
        except KeyError:
            val_b = 0
        try:
            val_c = c[key]
        except KeyError:
            val_c = 0
        f_data[key] = (val_b, val_c)
print f_data

    When I run this code, I get the following output:

    OrderedDict([('04-25', (5, 0)), ('04-23', (53, 0)), ('04-20', (3, 5)), ('04-2', (33, 0)), ('03-18', (0, 53)), ('03-13', (0, 3)), ('03-12', (0, 133))])