Python count same keys in ordered dict

Question

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":
        print(item["rel"])

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.

Your question is not clear. If there are no dicts with the key type as "a1", nothing will be printed... — Tomerikoo, Mar 14 '23 at 14:31
[Group/Count list of dictionaries based on value](https://stackoverflow.com/q/15815976/6045800) — Tomerikoo, Mar 14 '23 at 15:10
You are right, sorry for missleading, I modified my question then I think it is clear now. — Darwick, Mar 14 '23 at 15:56
I don't understand. You get the error because you treat your input (`d`) as a list of dicts, but it is actually a dict. Then you say you want to check *"if the key "type" has more then 0 value with "a1""*. What you really want to check is whether `d` is a list or a dict? — Tomerikoo, Mar 14 '23 at 15:59
Exactly. Now I realized the API answers, this would be the best. — Darwick, Mar 14 '23 at 16:10
[What's the canonical way to check for type in Python?](https://stackoverflow.com/q/152580/6045800) — Tomerikoo, Mar 14 '23 at 16:12

Kurt · Accepted Answer · 2023-03-14T16:21:23.940

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":
            print(item["rel"])
elif isinstance(d, OrderedDict):
    if d["type"] == "a1":
        print(d["rel"])

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.

score 0 · Answer 2 · answered Mar 14 '23 at 14:52

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
        print(item["rel"])

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
        print(item["rel"])

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"]

score -1 · Answer 3 · answered Mar 14 '23 at 14:58

-1

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:
    print(item["rel"])

answered Mar 14 '23 at 14:58

John Gordon

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That's not what is asked for. The question is not super clear, but it seems the key `type` is always there. OP is asking about when it has `a1` as the value – Tomerikoo Mar 14 '23 at 15:06

Python count same keys in ordered dict

3 Answers3