py2neo: Graph.find_one with multiple key/values

Question

I have some trouble with the py2neo find and find_one (http://py2neo.org/2.0/essentials.html)

What I want in Cypher is:

MATCH (p:Person) WHERE p.name='Alice' AND p.age=22 RETURN p

Say, where there are more than one key/value set (eg. if there are more than one 'Alice' in the graph).

My problem is that I don't know what to give graph.find_one, a working code is:

graph.find_one('Person', 'name', 'Alice')

What I would like is something like (This is not working!):

graph.find_one('Person', {'name': 'Alice', 'age': 22}) 

A possible (bad) solution would be to make a graph.find, and then loop through the results properties and look for the age, but I don't like that solution.

Bonus: Would it be possible with graph.find to do something like age > 25?

---

EDIT: New "solution"

find_person = "MATCH (p:Person) WHERE p.name = {N} AND p.age = {A} RETURN p"

>>> tx = graph.cypher.begin()
>>> tx.append(find_person, {'N': 'Alice', 'A': 22})
>>> res = tx.process()
>>> print(res[0][0][0])
(n423:Person {age:22,name:"Lisa"})

What I don't like about this is I miss the Note-object, (And I don't fully understand the RecordListList, and how to navigate it nicley)

Answer 1

Based on @elyase answer and the original py2neo.Graph.find, I've made this code. Please feel free to comment and improve.. :-)

def find_dict(graph, label, key_value=None, limit=None):
    """ Iterate through a set of labelled nodes, optionally filtering
    by property key/value dictionary
    """
    if not label:
        raise ValueError("Empty label")
    from py2neo.cypher.lang import cypher_escape
    if key_value is None:
        statement = "MATCH (n:%s) RETURN n,labels(n)" % cypher_escape(label)
    else:
        # quote string values
        d = {k: "'{}'".format(v) if isinstance(v, str) else v
             for k, v in key_value.items()}

        cond = ""
        for prop, value in d.items():
            if not isinstance(value, tuple):
                value = ('=', value)

            if cond == "":
                cond += "n.{prop}{value[0]}{value[1]}".format(
                    prop=prop,
                    value=value,
                )
            else:
                cond += " AND n.{prop}{value[0]}{value[1]}".format(
                    prop=prop,
                    value=value,
                )

        statement = "MATCH (n:%s ) WHERE %s RETURN n,labels(n)" % (
            cypher_escape(label), cond)
    if limit:
        statement += " LIMIT %s" % limit
    response = graph.cypher.post(statement)
    for record in response.content["data"]:
        dehydrated = record[0]
        dehydrated.setdefault("metadata", {})["labels"] = record[1]
        yield graph.hydrate(dehydrated)
    response.close()


def find_dict_one(graph, label, key_value=None):
    """ Find a single node by label and optional property. This method is
    intended to be used with a unique constraint and does not fail if more
    than one matching node is found.
    """
    for node in find_dict(graph, label, key_value, limit=1):
        return node

use of find_dict_one:

>>> a = find_dict_one(graph, 'Person', {'name': 'Lisa', 'age': 23})
>>>     print(a)
(n1:Person {age:23,name:"Lisa"})

Use of find_dict with tuple:

>>> a = find_dict(graph, 'Person', {'age': ('>', 21)}, 2)    >>> for i in a:
>>>     print(i)
(n2:Person {age:22,name:"Bart"})
(n1:Person {age:23,name:"Lisa"})

Use of find_dict without tuple:

>>> a = find_dict(graph, 'Person', {'age': 22}, 2)    >>> for i in a:
>>>     print(i)
(n2:Person {age:22,name:"Bart"})