Python count same keys in ordered dict


I have multiple ordered dict (in this example 3, but can be more) as a result from an API, with same key value pairs and I test for the same key, then I print out another key’s value:

from collections import OrderedDict

d = [OrderedDict([('type', 'a1'), ('rel', 'asd1')]),
     OrderedDict([('type', 'a1'), ('rel', 'fdfd')]),
     OrderedDict([('type', 'b1'), ('rel', 'dfdff')])]

for item in d:
    if item["type"] == "a1":

This works fine, but it could happen that the return from API can be different, because there is only 1 result. Without the [ at the begining and ] at the end, like this:

d = OrderedDict([('type', 'b1'), ('rel', 'dfdff')])

In this case, I will receive this error:

    if item["type"] == "a1":
TypeError: string indices must be integers

So I would like to test also (maybe in the same if it is possible) if the key "type" has more then 0 value with "a1" then print the items or to test the OrderedDict have the square brackets at the begining or at the end.

Asked By: Darwick



You can simply keep a flag that remembers if item["type"] == "a1" was ever true.

has_a1 = False
for item in d:
    if item["type"] == "a1":
        has_a1 = True

If you want the actual count, use an integer variable instead of a Boolean.

a1_count = 0
for item in d:
    if item["type"] == "a1":
        a1_count += 1

If you don’t mind using more memory (and maybe there’s a reason to do this anyway), keep a list of item["rel"] values instead of printing them on demand, then check the length of the list later.

a1_values = [item["rel"] for item in d if item["type"] == "a1"]
Answered By: chepner

If you want to check that a key is present in a dictionary, without caring what specific value it has, use this:

if key in mydict:

So in your case, it would look like this:

if "type" in item:
Answered By: John Gordon

The "square brackets" represent a list object. You simply need to check if d is a list or an OrderedDict and behave accordingly.

if isinstance(d, list):
    for item in d:
        if item["type"] == "a1":
elif isinstance(d, OrderedDict):
    if d["type"] == "a1":

The TypeError happens because when d is an OrderedDict rather than a list of OrderedDicts, the for loop iterates over the items in the OrderedDict, and those are strings.

Answered By: Kurt
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