I would like to compare between two list using item of the first list with index of the second list, and new list will append from second list for each matched.
a = [[1],[0],[0]]
b = [[1,2],[3,4],[5,6]]
c = []
for item in a:
for i in range(len(b)):
if item == b[i]:
c.append(b[i])
the answer should like this:
c = [[3,4],[1,2],[1,2]]
Simplest:
c = [b[i[0]] for i in a]
I recommend adding range checks though:
c = [b[i[0]] for i in a if (0 <= i[0] < len(b))]
EDIT: Based on your clarification of a, I recommend changing:
def findInstances(list1, list2):
for i in list1:
yield [pos for pos,j in enumerate(list2) if i==j] # This yields a list containing the value you want
to:
def findInstances(list1, list2):
for i in list1:
if (i in list2):
yield list2.index(i) # This yields only the value you want
This will flatten your list, and make the problem simpler. You can then use the following:
c = [b[i] for i in a if (0 <= i < len(b))]
Based on how you're getting a, the range checks are actually unnecessary. I left them in though, just in case you ever get a in a different way.
In [1]: a = [[1],[0],[0]]
In [2]: b = [[1,2],[3,4],[5,6]]
In [3]: [b[x[0]] for x in a]
Out[3]: [[3, 4], [1, 2], [1, 2]]
[b[x[0]] for x in a]
using numpy indexing:
>>a = np.asarray(a)
>>b = np.asarray(b)
>>b[a]
array([[[3, 4]],
[[1, 2]],
[[1, 2]]])
Your algorithm is almost correct. The problem is with the if statement. If you tried print out item and b[i] before testing for equality you would see the problem.
>>> a = [[1],[0],[0]]
>>> b = [[1,2],[3,4],[5,6]]
>>> c = []
>>> for item in a:
>>> for i in range(len(b)):
>>> print("item == b[i] is {} == {} is {}".format(item, b[i],
item == b[i]))
>>> if item == b[i]:
>>> c.append(b[i])
item == b[i] is [1] == [1, 2] is False
item == b[i] is [1] == [3, 4] is False
item == b[i] is [1] == [5, 6] is False
item == b[i] is [0] == [1, 2] is False
item == b[i] is [0] == [3, 4] is False
item == b[i] is [0] == [5, 6] is False
item == b[i] is [0] == [1, 2] is False
item == b[i] is [0] == [3, 4] is False
item == b[i] is [0] == [5, 6] is False
You have essentially been checking that each element in a and b for equality. Rather you want to check the elements in each item of a for equality with the index of b.
eg.
for item_a in a:
for index_b, item_b in enumerate(b):
# only check index 0 of item_a as all lists are of length one.
print("item_a[0] == index_b is {} == {} is {}".format(item_a[0],
index_b, item_a[0] == index_b))
if item_a[0] == index_b:
c.append(item_b)
produces:
item_a[0] == index_b is 1 == 0 is False
item_a[0] == index_b is 1 == 1 is True
item_a[0] == index_b is 1 == 2 is False
item_a[0] == index_b is 0 == 0 is True
item_a[0] == index_b is 0 == 1 is False
item_a[0] == index_b is 0 == 2 is False
item_a[0] == index_b is 0 == 0 is True
item_a[0] == index_b is 0 == 1 is False
item_a[0] == index_b is 0 == 2 is False
enumerate is a builtin helper function that returns a tuple containing the index and element for each element in a list (or anything that is iterable).
Unless you need to I would also recommend flattening a as having nested lists is redundant here ie. a = [1, 0, 0].
Having said all this, if you can get your head around list comprehensions then coding a solution would be much simpler -- as evidenced by the other answers to your question.
this is how i comparing between another two different list.
def findInstances(list1, list2):
for i in list1:
yield [pos for pos,j in enumerate(list2) if i==j]
list1 = [0.1408, 0.1456, 0.2118, 0.2521, 0.1408, 0.2118]
list2 = [0.1408, 0.1456, 0.2118, 0.2521, 0.1254, 0.1243]
list3 = [[1,2],[3,4],[5,6],[7,8],[9,10],[11,12]]
res = list(findInstances(list1, list2))
and res produce the output as 'a' in the first question
thank you
bby an index from lista