清理长且难看的if / elif / elif / elif ...块

Question

主持人注意

这个问题根本不同于所提到的可能重复的问题。这是关于清理一个长条件的并寻找一种更优雅的表达方式。

问题

这段代码很难看，不是pythonic。我是python的新手，不是编程的新手。不用猜测许多可能的伪装之一，您能告诉我执行此操作的pythonic方法吗？

如您所见，一堆变量（向前，向左，向右等）设置了浮点数。我想在变量“ closest_dir”中存储包含最小值的变量的字符串名称。

forward = calc_range(msg.ranges, 359, 0, 15)
right = calc_range(msg.ranges, 270, 271, 15)
left = calc_range(msg.ranges, 90, 91, 15)
back = calc_range(msg.ranges, 180, 181, 15)
narrow_l1 = sum(msg.ranges[83:87])/5
narrow_l2 = sum(msg.ranges[88:92])/5
narrow_l3 = sum(msg.ranges[93:97])/5
narrow_r1 = sum(msg.ranges[273:277])/5
narrow_r2 = sum(msg.ranges[268:272])/5
narrow_r3 = sum(msg.ranges[263:267])/5
closest_dist = min(narrow_l1, narrow_l2, narrow_l3, narrow_r1, narrow_r2, narrow_r3,
                   forward, left, right, back)
if (closest_dist == forward):
    closest_dir = "forward"
elif (closest_dist == left):
    closest_dir = "left"
elif (closest_dist == right):
    closest_dir = "right"
elif (closest_dist == back):
    closest_dir = "back"
elif (closest_dist == narrow_l1):
    closest_dir = "narrow_l1"
elif (closest_dist == narrow_l2):
    closest_dir = "narrow_l2"
elif (closest_dist == narrow_l3):
    closest_dir = "narrow_l3"
elif (closest_dist == narrow_r1):
    closest_dir = "narrow_r1"
elif (closest_dist == narrow_r2):
    closest_dir = "narrow_r2"
elif (closest_dist == narrow_r3):
    closest_dir = "narrow_r3"
else:
    closest_dir = "bug"

Answer 1

使用字典。

foos = {
    'forward': calc_range(msg.ranges, 359, 0, 15),
    'right': calc_range(msg.ranges, 270, 271, 15),
    'left': calc_range(msg.ranges, 90, 91, 15),
    'back': calc_range(msg.ranges, 180, 181, 15),
    'narrow_l1': sum(msg.ranges[83:87]/5),
    'narrow_l2': sum(msg.ranges[88:92])/5,
    'narrow_l3': sum(msg.ranges[93:97])/5,
    'narrow_r1': sum(msg.ranges[273:277])/5,
    'narrow_r2': sum(msg.ranges[268:272])/5,
    'narrow_r3': sum(msg.ranges[263:267])/5,
}

closest_dir = min(foos, key=foos.get)

min使用dict迭代器，该迭代器返回密钥，并且key自变量表示foo.get(x)将确定每个密钥{{1}的大小” }进行比较。