None of the other answers mention the primary reason for the difference in speed, which is that the zipped version avoids 10,000 tuple allocations. As a couple of the other answers do note, the zip version involves an intermediate array, while the zipped version doesn’t, but allocating an array for 10,000 elements isn’t what makes … Read more
This can be done using numpy.core.defchararray.add. Here is an example: >>> import numpy as np >>> a1 = np.array([‘a’, ‘b’]) >>> a2 = np.array([‘E’, ‘F’]) >>> np.core.defchararray.add(a1, a2) array([‘aE’, ‘bF’], dtype=”<U2″) There are other useful string operations available for NumPy data types.
Try a list comprehension: l = [x * 2 for x in l] This goes through l, multiplying each element by two. Of course, there’s more than one way to do it. If you’re into lambda functions and map, you can even do l = map(lambda x: x * 2, l) to apply the function … Read more
For elementwise multiplication of matrix objects, you can use numpy.multiply: import numpy as np a = np.array([[1,2],[3,4]]) b = np.array([[5,6],[7,8]]) np.multiply(a,b) Result array([[ 5, 12], [21, 32]]) However, you should really use array instead of matrix. matrix objects have all sorts of horrible incompatibilities with regular ndarrays. With ndarrays, you can just use * for … Read more
Use a list comprehension mixed with zip():. [a*b for a,b in zip(lista,listb)]
Use map with operator.add: >>> from operator import add >>> list( map(add, list1, list2) ) [5, 7, 9] or zip with a list comprehension: >>> [sum(x) for x in zip(list1, list2)] [5, 7, 9] Timing comparisons: >>> list2 = [4, 5, 6]*10**5 >>> list1 = [1, 2, 3]*10**5 >>> %timeit from operator import add;map(add, list1, … Read more
(A==B).all() test if all values of array (A==B) are True. Note: maybe you also want to test A and B shape, such as A.shape == B.shape Special cases and alternatives (from dbaupp’s answer and yoavram’s comment) It should be noted that: this solution can have a strange behavior in a particular case: if either A … Read more