如何在python中枚举OrderedDict

Question

So, I'm a php programmer who is trying to learn python. i have a dict of dict that i want sorted. I turned them into OrderedDict. They sort perfectly, The original dict look like this. This is just a 3 dimensional array right?

a["01/01/2001"]["un"]=1
a["01/01/2001"]["nn"]=1
a["01/02/2001"]["aa"]=2
a["01/02/2001"]["bb"]=2
a["01/03/2001"]["zz"]=3
a["01/03/2001"]["rr"]=3

I can convert them into OrderedDict, and want to present them in the following format

"01/01/2001" un=1 nn=1
"01/02/2001" aa=2 bb=2
"01/03/2001" zz=3 rr=3

I can write a simple loop in php to go through this associative array, but i can't figure out how to do it in python. Could someone help?

Answer 1

Loop through the keys and values using the dict.items() or dict.iteritems() methods; the latter lets you iterate without building an intermediary list of key-value pairs:

for date, data in a.iteritems():
    print date,
    for key, value in data.iteritems():
        print '{}={}'.format(key, value),
    print

Looping directly over dictionaries gives you keys instead; you can still access the values by using subscription:

for date in a:
    print date,
    for key in a[date]:
        print '{}={}'.format(key, a[date][key]),
    print

Answer 2

I think rather than OrderedDict, you will be better off with a defaultdict:

from collections import defaultdict

a = defaultdict(dict)
a["01/03/2001"]["zz"]=3
a["01/01/2001"]["un"]=1
a["01/02/2001"]["aa"]=2
a["01/01/2001"]["nn"]=1
a["01/02/2001"]["bb"]=2
a["01/03/2001"]["rr"]=3

# a is now a dict of dicts, each key is a date and each value is a dict of all 
# subkey-values

# print out in date order
for k,v in sorted(a.items()):
    # for each subdict, print key=value in sorted key order
    print k, ' '.join("%s=%s" % (kk,vv) for kk,vv in sorted(v.items()))

Prints:

01/01/2001 nn=1 un=1
01/02/2001 aa=2 bb=2
01/03/2001 rr=3 zz=3

EDIT:

Ah! My bad, you want the k=v values shown in insertion order, so you need a defaultdict of OrderedDict's:

from collections import defaultdict, OrderedDict

a = defaultdict(OrderedDict)
a["01/01/2001"]["un"]=1
a["01/01/2001"]["nn"]=1
a["01/02/2001"]["aa"]=2
a["01/02/2001"]["bb"]=2
a["01/03/2001"]["zz"]=3
a["01/03/2001"]["rr"]=3

# print out in date order
for k,v in sorted(a.items()):
    # for each subdict, print key=value in as-inserted key order, so no sort requred
    print k, ' '.join("%s=%s" % (kk,vv) for kk,vv in v.items())

Prints:

01/01/2001 un=1 nn=1
01/02/2001 aa=2 bb=2
01/03/2001 zz=3 rr=3

Answer 3

Try This Code.

a = {
        "01/01/2001": {"un": 1, "nn": 1}, \
        "01/02/2001": {"aa": 2, "bb": 2}, \
        "01/03/2001": {"zz": 3, "rr": 3}
    }

class decor(object):
    def __init__(self, a):
        self.a = a

    def __call__(self, fun):
        def wrapper(key, _list):
            print key, ' '.join(["{}={}".format(ele, self.a[key][ele]) \
                        for ele in _list
                    ])
            fun(key)
        return wrapper

@decor(a)
def main_fun(key):
    pass

for key in sorted(a.keys()):
    main_fun(key, [inner_key for inner_key in sorted(a[key].keys())])