To complete @Jeff's answer, we can compare the time of computation for the two methods:
import numpy as np import time test = np.random.randint(1, 50000, 10000) train = np.random.randint(1, 50000, 10000) start_list = time.time() overlap = [e for e in test if e in train ] end_list = time.time() print("with list comprehension: " + str(end_list - start_list)) set_test = set(test) set_train = set(train) start_set = time.time() overlap = set_test.intersection(set_train) end_set = time.time() print("with sets: " + str(end_set - start_set))
We get the output:
with list comprehension: 0.08894968032836914 with sets: 0.0003533363342285156
There are, however, specific methods to determine the intersection and the union of two sets. As long as ordering isn't important*, here's how you might do it:
train_set = set(train) # Use frozenset if no mutation is required test_set = set(test) common_elements = train_set & test_set # or, equivalently common_elements = train_set.intersection(test_set)
set_test = set(e) set_train = set(train) overlap = set_test.intersection(set_train)
You can use
import random import numpy as np train = [random.randint(1, 51) for var in range(1, 9000) ] #Your list test = [random.randint(1, 51) for var in range(1, 9000) ] #Your list train = np.array(train) #Converting list into numpy 's array test = np.array(test) overlap = np.intersect1d(train, test) print(overlap)
Last Updated : 01 Sep, 2021,GATE CS 2021 Syllabus
Input: lst1 = [15, 9, 10, 56, 23, 78, 5, 4, 9] lst2 = [9, 4, 5, 36, 47, 26, 10, 45, 87] Output: [9, 10, 4, 5] Input: lst1 = [4, 9, 1, 17, 11, 26, 28, 54, 69] lst2 = [9, 9, 74, 21, 45, 11, 63, 28, 26] Output: [9, 11, 26, 28]
[9, 11, 26, 28]
Working: The filter part takes each sublist’s item and checks to see if it is in the source list. The list comprehension is executed for each sublist in list2.
[ [13, 32], [7, 13, 28], [1, 6] ]
Objects of different types, except different numeric types, never compare equal. The == operator is always defined but for some object types (for example, class objects) is equivalent to is. The <, <=, > and >= operators are only defined where they make sense; for example, they raise a TypeError exception when one of the arguments is a complex number.,There are really two flavors of function objects: built-in functions and user-defined functions. Both support the same operation (to call the function), but the implementation is different, hence the different object types.,All numeric types (except complex) support the following operations (for priorities of the operations, see Operator precedence):,Python defines several iterator objects to support iteration over general and specific sequence types, dictionaries, and other more specialized forms. The specific types are not important beyond their implementation of the iterator protocol.
>>> n = -37 >>> bin(n) '-0b100101' >>> n.bit_length() 6
def bit_length(self): s = bin(self) # binary representation: bin(-37) -- > '-0b100101' s = s.lstrip('-0b') # remove leading zeros and minus sign return len(s) # len('100101') -- > 6
>>> n = 19 >>> bin(n) '0b10011' >>> n.bit_count() 3 >>> (-n).bit_count() 3
def bit_count(self): return bin(self).count("1")
>>> (1024).to_bytes(2, byteorder = 'big') b '\x04\x00' >>> (1024).to_bytes(10, byteorder = 'big') b '\x00\x00\x00\x00\x00\x00\x00\x00\x04\x00' >>> (-1024).to_bytes(10, byteorder = 'big', signed = True) b '\xff\xff\xff\xff\xff\xff\xff\xff\xfc\x00' >>> x = 1000 >>> x.to_bytes((x.bit_length() + 7) // 8, byteorder='little') b '\xe8\x03'
>>> int.from_bytes(b '\x00\x10', byteorder = 'big') 16 >>> int.from_bytes(b '\x00\x10', byteorder = 'little') 4096 >>> int.from_bytes(b '\xfc\x00', byteorder = 'big', signed = True) - 1024 >>> int.from_bytes(b '\xfc\x00', byteorder = 'big', signed = False) 64512 >>> int.from_bytes([255, 0, 0], byteorder = 'big') 16711680