Python: Test AB and CD are orthogonal or not
Python Basic - 1: Exercise-55 with Solution
There are four different points on a plane, P(xp,yp), Q(xq, yq), R(xr, yr) and S(xs, ys). Write a Python program to test AB and CD are orthogonal or not.
xp,yp, xq, yq, xr, yr, xs and ys are -100 to 100 respectively and each value can be up to 5 digits after the decimal point It is given as a real number including the number of. Output:
Output AB and CD are not orthogonal! or AB and CD are orthogonal!.
while True: try: print("Input xp, yp, xq, yq, xr, yr, xs, ys:") x_p, y_p, x_q, y_q, x_r, y_r, x_s, y_s = map(float, input().split()) pq_x, pq_y = x_q - x_p, y_q - y_p rs_x, rs_y = x_s - x_r, y_s - y_r rs = pq_x*rs_x + pq_y*rs_y if abs(rs) > 1e-10: print("AB and CD are not orthogonal!") else: print("AB and CD are orthogonal!") except: break
Input xp, yp, xq, yq, xr, yr, xs, ys: 4.5 -2.5 -2.5 4.5 3.5 3.5 3.8 -3.5 AB and CD are not orthogonal!
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Python: Tips of the Day
How to make a flat list out of list of lists?
Given a list of lists l
flat_list = [item for sublist in l for item in sublist]
flat_list =  for sublist in l: for item in sublist: flat_list.append(item)
is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:
flatten = lambda l: [item for sublist in l for item in sublist]
As evidence, you can use the timeit module in the standard library:
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' '[item for sublist in l for item in sublist]' 10000 loops, best of 3: 143 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'sum(l, )' 1000 loops, best of 3: 969 usec per loop $ python -mtimeit -s'l=[[1,2,3],[4,5,6], , [8,9]]*99' 'reduce(lambda x,y: x+y,l)' 1000 loops, best of 3: 1.1 msec per loop
Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.
The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.
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