从字典中查找可用数量

问题描述

我写了一个逻辑，以查找位置上的可用数量，位置和数量通过字典进行管理，

I have written a logic to find available quantity in location,for the location and quantity is managed with dictionary,

d={'loc2': 500.0, 'loc3': 200.0, 'loc1': 1000.0, 'loc4': 100.0, 'loc5': 50.0}

def find_combination(locations,qty):
    new_list = sorted(locations.items(),key=lambda y: y[1],reverse=True)
    result = []
    while qty > 0:
        min_item = ''
        for item in new_list:
            if item[0] in result:
                continue
            new_diff = abs(qty - item[1])
            if not min_item or new_diff <= min_diff:
                min_item = item[0]
                min_diff = new_diff
                min_val = item[1]
        result.append((min_item ,locations.get(min_item)))
        qty = qty - min_val
    return result

现在，当数量小于字典中的最大数量时，它会产生意外的结果，

Now when the quantity is nelow the max qty in the dict it's giving unexpected result,

print find_combination(d,500)
OUTPUT: [('loc2', 500.0)]
print find_combination(d,1000)
OUTPUT: [('loc1', 1000.0)]
print find_combination(d,750)
OUTPUT: [('loc2', 500.0), ('loc3', 200.0), ('loc5', 50.0)]
print find_combination(d,1800)
OUTPUT: [('loc1', 1000.0), ('loc1', 1000.0)] # unexpected

推荐答案

您能解释一下为什么该输出是意外的吗?将loc1项附加到result后，qty的值将为800.在下一次迭代中，行new_diff = abs(qty - item[1])将再次为项loc1返回最小值(200)，因此该项将再次添加到result中.完成此操作后，qty将是-200，因此while循环将终止.仅当相关数量小于变量qty时才添加项目吗?如果是这样，您需要更多的逻辑来做到这一点-您可以将for循环更改为:

Could you explain why that output is unexpected? After one loc1 item has been appended to result, the value of qty will be 800. The line new_diff = abs(qty - item[1]) will return a minimal value (200) for the item loc1 again on the next iteration, so that item will be added to result once again. Once that's been done, qty will be -200, so the while loop will terminate. Should you only be adding items if their associated quantity is smaller than the variable qty? If so, you need more logic to do that - you could change the for loop to:

for item in [x for x in new_list if x[1] <= qty]:

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