问题
I have the following sorted python list, although multiple values can occur:
[0.0943200769115388, 0.17380131294164516, 0.4063245853719435,
0.45796523225774904, 0.5040225609708342, 0.5229351852840304,
0.6145136350368882, 0.6220712583558284, 0.7190096076050408,
0.8486436998476048, 0.8957381707345986, 0.9774325873910711,
0.9832076130275351, 0.985386554764682, 1.0]
Now, I want to know the index in the array where a particular value may fall:
For example, a value of 0.25 would fall in index 2 because it is between 0.173 and 0.40. I guess I can go through the list and do this in a for loop but I was wondering if there is some better way to do this which maybe more computationally efficient. I create this array once but have to perform many lookups.
回答1:
>>> vals = [0.0943200769115388, 0.17380131294164516, 0.4063245853719435,
0.45796523225774904, 0.5040225609708342, 0.5229351852840304,
0.6145136350368882, 0.6220712583558284, 0.7190096076050408,
0.8486436998476048, 0.8957381707345986, 0.9774325873910711,
0.9832076130275351, 0.985386554764682, 1.0]
>>> import bisect
>>> bisect.bisect(vals, 0.25)
2
回答2:
If you know the list is already sorted, then the textbook solution is to do a binary search. You keep two index bounds, min and max. Initialize them to 0 and len - 1. Then set mid to be (min + max) / 2. Compare the value at index mid with your target value. If it's less, then set min to mid + 1. If it's greater, then set max to mid - 1. Repeat until you either find the value or until max < min, in which case you will have found the desired index in O(log(n)) steps.
来源：https://stackoverflow.com/questions/33323892/python-find-value-within-range-in-float-array