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For a list, sorting is very simple. To sort in forward order (from small to large) use
list.sort()
and to sort in reverse order (from large to small) use
list.sort(reverse=True)
. But if you are not limited to getting a sorted list, you also want to record the original next mark, then for a numpy.array, you can use
np.argsort()
For example, [1,3,2,5,6]
will become [1,2,3,5 after sorting ,6]
But if we want to know what the original subscript of the sorted result is (the answer is [0,2,1,3,4]), we can use np.argsort()
But if it is a simple list and you want to achieve this effect, you can use
# enumerate(x)会自动构造一个tuple(a,b) # 其中a是index,b是list里index下标对应的具体的值,后面的x是代表一个虚拟变量,即tuple(a,b) sorted_list = sorted(enumerate(list), key=lambda x:x[1]) # x[1]是因为在enumerate(a)中,a数值在第1位 result = [x[0] for x in sorted_list]
The result returned is the original index
If you want to achieve the one in the title, go one step further , that is, to get a new subscript after sorting the list, for example, for [1,5,2,8,3,4], you should get [0,4,1,5,2,3]
Then you can use
# enumerate(x)会自动构造一个tuple(a,b) # 其中a是index,b是list里index下标对应的具体的值,后面的x是代表一个虚拟变量,即tuple(a,b) # sorted_list = [(0,1),(2,2),(4,3),(5,4),(1,5),(3,8)] sorted_list = sorted(enumerate(list), key=lambda x:x[1]) # x[1]是因为在enumerate(a)中,a数值在第1位 for i in range(len(sorted_list)): list[sorted_list[i][1]] = i
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