清单中项目的处理

Question

Consider I have Dictionary named "tenant_id_dict" with entry as follows:

defaultdict(<type 'list'>, {u'b0116ce25cad4106becbbddfffa61a1c': [u'demo_ins1'], u'1578f81703ec4bbaa1d548532c922ab9': [u'new_ins_1', u'new_tenant_ins']})

That is combination of tenant_id and instance_name.

Now I need to utilize the tenant_id and instance_name as needed.

Consider the following code:

for tenants,instances in tenant_id_dict.iteritems():

        compute_value_for_instance = ck.reports.get_total(tenant_id=tenants, service='compute', instance_name=instances)
        print compute_value_for_instance

Here I need to process the values for tenants and instance_name dynamically.

But In above Code I am unable to get the result correctly as Instance is having the list of values.

I need to process till the last item in a list.

Someone let me know the way to the processing of values here.

Answer 1

You initialized a defaultdit with list . This list is iterable and value for each key. So you need to iterate over the values (ie list ) as below.

for tenants,instances in tenant_id_dict.iteritems():
        for single_ins in instances:
                compute_value_for_instance = ck.reports.get_total(tenant_id=tenants, service='compute', instance_name=single_ins)
                print compute_value_for_instance

Working Demo-

from collections import defaultdict
#Create a dummy defaultdict with list values
d = defaultdict(list)

for i in range(5):
    d[i].append(i+100)
    d[i].append(i+200)
    d[i].append(i+300)


#Experiment
for k,v in d.iteritems():
    for single_vale in v:#Here v is a list
        print str(k)+str(single_vale)
print d

Output-

0100
0200
0300
1101
1201
1301
2102
2202
2302
3103
3203
3303
4104
4204
4304
defaultdict(<type 'list'>, {0: [100, 200, 300], 1: [101, 201, 301], 2: [102, 202, 302], 3: [103, 203, 303], 4: [104, 204, 304]})