'dict' object has no attribute 'has_key'

Question

While traversing a graph in Python, a I'm receiving this error:

'dict' object has no attribute 'has_key'

Here is my code:

def find_path(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return path
    if not graph.has_key(start):
        return None
    for node in graph[start]:
        if node not in path:
            newpath = find_path(graph, node, end, path)
            if newpath: return newpath
    return None

The code aims to find the paths from one node to others. Code source: http://cs.mwsu.edu/~terry/courses/4883/lectures/graphs.html

Why am I getting this error and how can I fix it?

Possible duplicate of 'has_key()' or 'in'?

Open AI - Opting Out
– Open AI - Opting Out

2015-11-16 06:22:20 +00:00
Commented Nov 16, 2015 at 6:22 — Open AI - Opting Out
– Open AI - Opting Out, Commented Nov 16, 2015 at 6:22

johnnyRose · Accepted Answer · 2017-12-13 15:13:12Z

329

has_key was removed in Python 3. From the documentation:

Removed dict.has_key() – use the in operator instead.

Here's an example:

if start not in graph:
    return None

edited Dec 13, 2017 at 15:13

answered Nov 16, 2015 at 1:23

johnnyRose

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4 Comments

Adam Smith Over a year ago

I think key not in d.keys() is probably much slower, too, since key not in d should be O(1) lookup and I believe keys produces a list, which is O(n) lookup (not to mention taking extra space in memory). I could be wrong about that though -- it might still be hashed lookup

Antti Haapala Over a year ago

@AdamSmith not in Python 3, d.keys() is a view that implements most of the set interface.

林果皞 Over a year ago

It removed... but why ? Since it make python 2 port to python 3 more work to do.

johnnyRose Over a year ago

@林果皞: The whole point of a new major version is that the developers can introduce improvements which may include breaking changes instead of having to support old features as the language matures. This is always a risk which must be considered before upgrading to a new major version. In this case, in is shorter and more Pythonic, as well as being consistent with other collections in the language.

qloveshmily · Accepted Answer · 2019-05-29 08:25:03Z

70

In python3, has_key(key) is replaced by __contains__(key)

Tested in python3.7:

a = {'a':1, 'b':2, 'c':3}
print(a.__contains__('a'))

answered May 29, 2019 at 8:25

qloveshmily

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2 Comments

KK2491 Over a year ago

I think this is the right and easier way to do, Thank you for the answer

jasonharper Over a year ago

No, no, no... the __contains__() method is how you implement a containment test in a user-defined class, the in operator is how you actually use that test. About the only time you should ever explicitly call a double-underscored "magic" method is in a subclass method, as part of calling the superclass implementation.

Abhishek Pansotra · Accepted Answer · 2018-08-24 10:04:23Z

34

has_key has been deprecated in Python 3.0. Alternatively you can use 'in'

graph={'A':['B','C'],
   'B':['C','D']}

print('A' in graph)
>> True

print('E' in graph)
>> False

answered Aug 24, 2018 at 10:04

Abhishek Pansotra

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Comments

Kevin S · Accepted Answer · 2015-11-16 02:52:50Z

9

I think it is considered "more pythonic" to just use in when determining if a key already exists, as in

if start not in graph:
    return None

answered Nov 16, 2015 at 2:52

Kevin S

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1 Comment

Amitay Drummer Over a year ago

I'm not sure, according to The Zen of Python(PEP 20): "Explicit is better than implicit". I think that if you use the in keyword, your intention might not be clear enough what does if start not in graph: means? may be graph is a list and it checks if there is no such string in the list? On the other hand, if you use syntax like has_key (now deprecated) or at least in graph.keys() it is more clear that graph is a dict

Godfrey · Accepted Answer · 2020-06-25 20:39:17Z

7

Try:

if start not in graph:

For more info see ProgrammerSought

answered Jun 25, 2020 at 20:39

Godfrey

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Comments

Oana Roxana · Accepted Answer · 2017-12-18 19:14:51Z

4

The whole code in the document will be:

graph = {'A': ['B', 'C'],
             'B': ['C', 'D'],
             'C': ['D'],
             'D': ['C'],
             'E': ['F'],
             'F': ['C']}
def find_path(graph, start, end, path=[]):
        path = path + [start]
        if start == end:
            return path
        if start not in graph:
            return None
        for node in graph[start]:
            if node not in path:
                newpath = find_path(graph, node, end, path)
                if newpath: return newpath
        return None

After writing it, save the document and press F 5

After that, the code you will run in the Python IDLE shell will be:

find_path(graph, 'A','D')

The answer you should receive in IDLE is

['A', 'B', 'C', 'D']

answered Dec 18, 2017 at 19:14

Oana Roxana

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1 Comment

Encipher Over a year ago

Can you please explain it ?specifically the recursion portion.

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