223 lines
5.7 KiB
Python
223 lines
5.7 KiB
Python
from collections.abc import Iterable
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from collections.abc import Iterator
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from itertools import chain
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from itertools import tee
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__all__ = [
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"split_len",
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"split",
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"nodes_equal",
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"edges_equal",
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"pairwise",
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"graphs_equal",
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# "arbitrary_element"
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]
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def split_len(nodes, step=30000):
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ret = []
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length = len(nodes)
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for i in range(0, length, step):
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ret.append(nodes[i : i + step])
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if len(ret[-1]) * 3 < step:
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ret[-2] = ret[-2] + ret[-1]
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ret = ret[:-1]
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return ret
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def split(nodes, n):
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ret = []
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length = len(nodes) # 总长
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step = int(length / n) + 1 # 每份的长度
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for i in range(0, length, step):
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ret.append(nodes[i : i + step])
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return ret
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def nodes_equal(nodes1, nodes2):
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"""Check if nodes are equal.
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Equality here means equal as Python objects.
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Node data must match if included.
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The order of nodes is not relevant.
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Parameters
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----------
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nodes1, nodes2 : iterables of nodes, or (node, datadict) tuples
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Returns
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-------
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bool
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True if nodes are equal, False otherwise.
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"""
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nlist1 = list(nodes1)
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nlist2 = list(nodes2)
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try:
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d1 = dict(nlist1)
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d2 = dict(nlist2)
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except (ValueError, TypeError):
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d1 = dict.fromkeys(nlist1)
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d2 = dict.fromkeys(nlist2)
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return d1 == d2
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def edges_equal(edges1, edges2, need_data=True):
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"""Check if edges are equal.
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Equality here means equal as Python objects.
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Edge data must match if included.
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The order of the edges is not relevant.
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Parameters
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----------
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edges1, edges2 : iterables of with u, v nodes as
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edge tuples (u, v), or
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edge tuples with data dicts (u, v, d), or
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edge tuples with keys and data dicts (u, v, k, d)
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Returns
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-------
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bool
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True if edges are equal, False otherwise.
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"""
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from collections import defaultdict
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d1 = defaultdict(dict)
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d2 = defaultdict(dict)
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c1 = 0
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for c1, e in enumerate(edges1):
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u, v = e[0], e[1]
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data = []
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if need_data == True:
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data = [e[2:]]
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if v in d1[u]:
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data = d1[u][v] + data
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d1[u][v] = data
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d1[v][u] = data
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c2 = 0
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for c2, e in enumerate(edges2):
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u, v = e[0], e[1]
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data = []
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if need_data == True:
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data = [e[2:]]
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if v in d2[u]:
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data = d2[u][v] + data
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d2[u][v] = data
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d2[v][u] = data
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if c1 != c2:
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return False
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# can check one direction because lengths are the same.
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for n, nbrdict in d1.items():
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for nbr, datalist in nbrdict.items():
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if n not in d2:
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return False
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if nbr not in d2[n]:
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return False
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d2datalist = d2[n][nbr]
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for data in datalist:
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if datalist.count(data) != d2datalist.count(data):
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return False
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return True
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# Recipe from the itertools documentation.
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def pairwise(iterable, cyclic=False):
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"s -> (s0, s1), (s1, s2), (s2, s3), ..."
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a, b = tee(iterable)
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first = next(b, None)
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if cyclic is True:
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return zip(a, chain(b, (first,)))
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return zip(a, b)
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def graphs_equal(graph1, graph2):
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"""Check if graphs are equal.
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Equality here means equal as Python objects (not isomorphism).
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Node, edge and graph data must match.
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Parameters
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----------
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graph1, graph2 : graph
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Returns
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-------
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bool
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True if graphs are equal, False otherwise.
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"""
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return (
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graph1.adj == graph2.adj
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and graph1.nodes == graph2.nodes
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and graph1.graph == graph2.graph
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)
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# def arbitrary_element(iterable):
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# """Returns an arbitrary element of `iterable` without removing it.
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# This is most useful for "peeking" at an arbitrary element of a set,
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# but can be used for any list, dictionary, etc., as well.
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# Parameters
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# ----------
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# iterable : `abc.collections.Iterable` instance
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# Any object that implements ``__iter__``, e.g. set, dict, list, tuple,
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# etc.
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# Returns
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# -------
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# The object that results from ``next(iter(iterable))``
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# Raises
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# ------
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# ValueError
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# If `iterable` is an iterator (because the current implementation of
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# this function would consume an element from the iterator).
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# Examples
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# --------
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# Arbitrary elements from common Iterable objects:
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# >>> eg.utils.arbitrary_element([1, 2, 3]) # list
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# 1
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# >>> eg.utils.arbitrary_element((1, 2, 3)) # tuple
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# 1
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# >>> eg.utils.arbitrary_element({1, 2, 3}) # set
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# 1
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# >>> d = {k: v for k, v in zip([1, 2, 3], [3, 2, 1])}
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# >>> eg.utils.arbitrary_element(d) # dict_keys
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# 1
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# >>> eg.utils.arbitrary_element(d.values()) # dict values
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# 3
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# `str` is also an Iterable:
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# >>> eg.utils.arbitrary_element("hello")
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# 'h'
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# :exc:`ValueError` is raised if `iterable` is an iterator:
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# >>> iterator = iter([1, 2, 3]) # Iterator, *not* Iterable
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# >>> eg.utils.arbitrary_element(iterator)
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# Traceback (most recent call last):
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# ...
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# ValueError: cannot return an arbitrary item from an iterator
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# Notes
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# -----
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# This function does not return a *random* element. If `iterable` is
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# ordered, sequential calls will return the same value::
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# >>> l = [1, 2, 3]
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# >>> eg.utils.arbitrary_element(l)
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# 1
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# >>> eg.utils.arbitrary_element(l)
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# 1
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# """
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# if isinstance(iterable, Iterator):
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# raise ValueError("cannot return an arbitrary item from an iterator")
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# # Another possible implementation is ``for x in iterable: return x``.
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# return next(iter(iterable))
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