如何在元組列表中找到第一次的最大絕對索引？

Question

給定一個元組列表：

[(-8.33958, -84.01769099999999), (-7.96129, -84.37399199999999), (-8.33958, -83.84126699999999), (-7.96129, -84.19756499999998), (-7.24097, -85.581651), (-6.86267, -85.937952), (-7.24097, -85.405227), (-6.86267, -85.76152499999999), (-9.4382, -83.473767), (-9.4382, -83.473767), (-8.35625, -85.11197399999999), (-9.0599, -83.83006499999999), (-9.0599, -83.83006499999999), (-7.97795, -85.46824499999998), (-8.35625, -84.93524699999999), (-7.97795, -85.29151799999998), (-8.33958, -85.03772699999999), (-8.00311, -88.17046199999997), (-5.88285, -86.59070369999999), (-6.86267, -88.655385), (-9.37174, -86.88101999999999), (-7.34506, -88.24291199999999), (-8.22317, -87.13259099999999), (-7.72054, -86.124309), (-8.35625, -86.131707), (-8.35625, -86.131707), (-7.03703, -88.90182), (-8.51394, -86.422992), (-8.51394, -85.83968999999999), (-9.61255, -85.510092), (-9.89311, -84.10327799999999), (-7.96129, -87.540312), (-9.13791, -86.022645)]

目的是找到元組中第一項的最高值的索引。 在輸入上方的示例輸入中，元組的索引(-9.89311, -84.10327799999999)

我一直在這樣做（但它不會返回正確的輸出）：

x = [(-8.33958, -84.01769099999999), (-7.96129, -84.37399199999999), (-8.33958, -83.84126699999999), (-7.96129, -84.19756499999998), (-7.24097, -85.581651), (-6.86267, -85.937952), (-7.24097, -85.405227), (-6.86267, -85.76152499999999), (-9.4382, -83.473767), (-9.4382, -83.473767), (-8.35625, -85.11197399999999), (-9.0599, -83.83006499999999), (-9.0599, -83.83006499999999), (-7.97795, -85.46824499999998), (-8.35625, -84.93524699999999), (-7.97795, -85.29151799999998), (-8.33958, -85.03772699999999), (-8.00311, -88.17046199999997), (-5.88285, -86.59070369999999), (-6.86267, -88.655385), (-9.37174, -86.88101999999999), (-7.34506, -88.24291199999999), (-8.22317, -87.13259099999999), (-7.72054, -86.124309), (-8.35625, -86.131707), (-8.35625, -86.131707), (-7.03703, -88.90182), (-8.51394, -86.422992), (-8.51394, -85.83968999999999), (-9.61255, -85.510092), (-9.89311, -84.10327799999999), (-7.96129, -87.540312), (-9.13791, -86.022645)]


index_of_max_abs_j = -1
for i, (j,k) in enumerate(x):
    if j*j > index_of_max_abs_j:
        index_of_max_abs_j = i
print index_of_max_abs_j

該代碼返回max(j*j)的索引，對嗎？ 與嘗試找到max(|j|)什么不同？

但是，還有另一種方法可以實現相同的輸出嗎？ 也許與sorted和reverse和一些math.abs key ？ 帶sorted ， reverse和/或key的替代方法是否更有效？

如果元組列表中有任何項具有相同的值，則返回最大絕對值的第一個實例的第一個索引。

Answer 1

我不確定為什么要將元組的第一個元素的值與索引進行比較，但這會為您獲取元組的索引，其中第一個元素的絕對值最高。

max_index = -1
max_value = 0
for i, z in enumerate(x):
    value = abs(z[0])
    if value > max_value:
        max_index = i
        max_value = value

print(x[max_index])

或作為不太易讀的單行代碼，

print(x.index(max(x, key=lambda y:abs(y[0]))))

Answer 2

這是一個替代方案，沒有對效率進行評論：

y = [abs(item[0]) for item in x]
m = max(y)
print(y.index(m))

Answer 3

print(max((abs(value[0]),index) for (index,value) in enumerate(values))[1]

將通過本機函數一次性給出具有max元素的最后一個元素的索引，因此應該是最快的方法

如果您真的想擁有最大數量的元素，則可以

print(-max((abs(value[0]),-index) for (index,value) in enumerate(values))[1])

Answer 4

您可以分2個步驟進行操作，以第1步中的第一個鍵為基礎獲取最大值，然后使用元素獲取索引

>>> max_value = sorted(x, key=lambda y: y[0])[0]

>>> x.index(max_value)

30