两个列表列表之间的公共元素(嵌套列表的交集)
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Common elements between two lists of lists (intersection of nested lists)
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我有两个大的二维点列表，如果有的话，我想找到它们的公共子列表。这两个列表都很大，效率是个问题。
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]我尝试了下面的 itertools，但我得到"ValueError：具有多个元素的数组的真值不明确。使用 a.any() 或 a.all()"。
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]我也从这里尝试了第一个答案，但我得到 'numpy.int64' object is not iterable。
另外，我认为这个简单的代码可以工作，但是需要很多时间：
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]有什么建议吗？谢谢。
列表是不可散列的，所以我们需要将内部列表转换为元组，然后我们可以使用集合交集来查找公共元素
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]输出
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]由于我们将列表分成集合，因此我们不考虑重复。考虑以下输入
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]我们在两个列表中都有两个 [3,41]，但是前面的 python 程序将在输出中只输出一个 [3,41]。以下程序将通过最初计算重复条目并在之后重复它们来处理重复条目。
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]输出
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]如果你真的只使用 list 那么，你可以从列表中创建 set 并为你的情况使用 set().intersection() -
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
for i in itertools.chain.from_iterable(t1):
if i in t2:
print"yes",iintersection = [i for i in t1 if i in t2]
t1 = [[3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[161, 160], [169, 260], [187, 540], [192, 10], [205, 23], [3,41]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
print st1.intersection(st2)
set([3,41])
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
t1 = [[3, 41], [3, 41], [5, 82], [10, 31], [11, 34], [14, 54]]
t2 = [[3,41], [3,41], [161, 160], [169, 260], [187, 540], [192, 10], [205, 23]]
nt1 = map(tuple, t1)
nt2 = map(tuple, t2)
st1 = set(nt1)
st2 = set(nt2)
from collections import defaultdict
d1 = defaultdict(int)
d2 = defaultdict(int)
for i in nt1:
d1[i] += 1#counting element occurrence from first list
for i in nt2:
d2[i] += 1 #counting element occurrence from second list
result_list = []
for i in st1.intersection(st2):
min_count = min(d1[i], d2[i]) #selecting the minimum one to multiply
result_list+=map(lambda x:list(i), xrange(0, min_count))
print result_list
[[3, 41], [3, 41]]
l1 = [[1,2],[2,3]]
l2 = [[3,4],[2,3]]
list(set(map(tuple,l1)).intersection(set(map(tuple,l2))))
>> [(2, 3)]但是对于非常非常大的 lists，这种方法可能会很慢。
编辑：使用 map 函数。