给定一个元组列表:
[(-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)]
目的是找到元组中第一项的最大值的索引。使用上面的示例输入，输出元组的索引 (-9.89311, -84.10327799999999)
我一直这样做(但它没有返回正确的输出):
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
代码返回 max(j*j) 的索引，但对吗？它与试图找到 max(|j|) 有什么不同吗？
但是还有另一种方法可以实现相同的输出吗？也许可以使用 sorted 和 reverse 以及 key 和一些 math.abs？使用 sorted、reverse 和/或 key 的替代方法是否更有效？
如果元组列表中有任何项具有相同的值，则返回最大绝对值的第一个实例的第一个索引。
请您参考如下方法:
我不确定您为什么要将元组第一个元素的值与索引进行比较，但这会为您提供第一个元素绝对值最高的元组索引。
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])
或者作为一种可读性较差的单行，
print(x.index(max(x, key=lambda y:abs(y[0]))))
