获取索引值的有效方法

Question

我有一个x，y值的元组列表。我想在列表中找到最接近的x值的索引。下面是我的代码

# list of coords
a = [(376, 220), (350, 218), (324, 216), (298, 214), (271, 211), (245, 210), (219, 208), (192, 205), (166, 204)]
to_find = (190, 210)

#grab a new list with only x axis elements
lst = []
for i in range(len(a)):
    lst.append(a[i][0])

#list of all x coordinates
print(lst)

#find the min closest element
def min_closest(lst, K):
    return lst[min(range(len(lst)), key=lambda i: abs(lst[i] - K))]

#print the corresponding index
print(lst.index(min_closest(lst, to_find[0])))

我用x值建立了一个新的列表。最后，我将搜索列表的x值与x列表进行比较，以找到最接近的元素。后来我抓起了它的索引。有什么有效的方法吗

Answer 1

将a转换为numpy.array，然后使用np.argmin：

arr = np.array(a)
diffs = np.abs(arr - to_find)
arr[np.argmin(diffs[:, 0])]
#OUTPUT array([192, 205])

Answer 2

尝试使用^{}：

from scipy.spatial.distance import euclidean
a = [(376, 220), (350, 218), (324, 216), (298, 214), (271, 211), (245, 210), (219, 208), (192, 205), (166, 204)]
to_find = (190, 210)
print(min(a, key = lambda x: euclidean(x, to_find)))

输出：

(192, 205)

Answer 3

你做了整件事，但又多了一步：

a = [(376, 220), (350, 218), (324, 216), (298, 214), (271, 211), (245, 210), (219, 208), (192, 205), (166, 204)]
to_find = (190, 210)

ix = min(range(len(a)), key = lambda x: abs(a[x][0] - to_find[0]))
print(ix)

输出：

7

另一种方法可能更快：

a = [(376, 220), (350, 218), (324, 216), (298, 214), (271, 211), (245, 210), (219, 208), (192, 205), (166, 204)]
to_find = (190, 210)

min_diff, min_ix = 999999999, None
for ix, value in enumerate(a):
    diff = abs(to_find[0] - value[0])
    if diff < min_diff:
        min_diff, min_ix = diff, ix
print(min_ix)