how do I build a for loop in order to print all float values in this nested dictionary, for any user?
plist = {'user1': {u'Fake Plastic Trees': 1.0, u'The Numbers': 1.0, u'Videotape': 1.0}}
desired output = [1.0, 1.0, 1.0]
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8-Bit Borges
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the values in the inner dictionary are already floats, what do you need? – taylor swift Nov 10 '16 at 17:48
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@data_garden: you didn't give us any expected output. Also, how much variation is there in your dictionary structure? Can there be an arbitrary level of nesting? – Martijn Pieters Nov 10 '16 at 17:49
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@data_garden: also, have you tried anything yourself yet? Did you get stuck anywhere? – Martijn Pieters Nov 10 '16 at 17:49
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Are there any *other* values in the dictionary that are *not* floats? – Martijn Pieters Nov 10 '16 at 17:50
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If not, this is as simple as `list(d['playlist'].values())` (where `list()` can be dropped in Python 2). – Martijn Pieters Nov 10 '16 at 17:50
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I cannot depend on 'playlist' as a constant `key` – 8-Bit Borges Nov 10 '16 at 17:51
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Did you search SO for *nested dictionaries*, I imagine there are many Q&A's here, one may fit your needs. – wwii Nov 10 '16 at 17:51
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@data_garden what *can* you depend on? This question isn't nearly specific enough at this point. – Martijn Pieters Nov 10 '16 at 17:52
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please refer to edit and see if that clarifies the quesiton – 8-Bit Borges Nov 10 '16 at 17:55
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@data_garden: sorry, it doesn't. Will there always be *one* key, or can there be multiple? What should happen if there are multiple? – Martijn Pieters Nov 10 '16 at 17:56
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@data_garden: you also didn't answer my question about what other values there might be in the nested dictionary. – Martijn Pieters Nov 10 '16 at 17:56
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always one `key`. – 8-Bit Borges Nov 10 '16 at 17:57
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http://stackoverflow.com/questions/10756427/loop-through-all-nested-dictionary-values may provide insight – wwii Nov 10 '16 at 18:03
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@ Martijn Pieters I had tried `[v.values() for v in plist.keys()]` – 8-Bit Borges Nov 10 '16 at 18:03
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@data_garden Is there a need to retrieve all of the values for multiple keys? Either together in one list or as a list of lists? – Taylor D. Edmiston Nov 10 '16 at 18:08
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if that would provide more versatility to the code, in one list, yes – 8-Bit Borges Nov 10 '16 at 18:09
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@data_garden I've updated my answer with one that supports multiple keys as well. – Taylor D. Edmiston Nov 10 '16 at 18:32
2 Answers
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There's a dict.values() method that does exactly what you want.
a_dict = {'user1': {u'Fake Plastic Trees': 1.0, u'The Numbers': 1.0, u'Videotape': 1.0}}
first_key = list(a_dict.keys())[0]
values = a_dict[first_key].values()
print(list(values))
Output
[1.0, 1.0, 1.0]
Edit: And if you want to return one flattened list of all values for all keys as mentioned in the comments on the question, you can do this:
a_dict = {
'user1': {u'Fake Plastic Trees': 1.0, u'The Numbers': 2.0, u'Videotape': 3.0},
'user2': {u'Foo': 4.0, u'Bar': 5.0},
}
values = []
for k in a_dict.keys():
for v in a_dict[k].values():
values.append(v)
print(values)
Output
[4.0, 5.0, 3.0, 1.0, 2.0]
Taylor D. Edmiston
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If something is incorrect about this answer with respect to your question, feel free to add a comment so it can be clarified. – Taylor D. Edmiston Nov 10 '16 at 17:57
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Im sorry, I haven't downvoted it. but I've edited the question so it must not depend on a specific `string` as the outer `key` – 8-Bit Borges Nov 10 '16 at 17:59
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@data_garden I've updated the answer to pull from the first key in the dictionary per the updated question. Note that this assumes there is only one key in the dictionary, which I believe is the case from reading your comments on the question. – Taylor D. Edmiston Nov 10 '16 at 18:04
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Use dict.values() to get your desired behavior.
>>> plist = {'playlist': {u'Fake Plastic Trees': 1.0, u'The Numbers': 1.0, u'Videotape': 1.0}}
>>> list(plist['playlist'].values())
[1.0, 1.0, 1.0]
>>>
Christian Dean
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