擅长:python、mysql、java
<p>显而易见的解决方案是:</p>
<pre><code>l = lambda r: 1 if r < 10 else 1.0-(r-9)/10.0 # 1 for r in 0..9, then less then 1 down to 0
value = 20
for r in range(20):
print (r,":", value * l(r)) # multiply your loss with round-based lambda factor
</code></pre>
<p>输出:</p>
<pre><code>0 : 20
1 : 20
2 : 20
3 : 20
4 : 20
5 : 20
6 : 20
7 : 20
8 : 20
9 : 20
10 : 18.0
11 : 16.0
12 : 14.0
13 : 12.0
14 : 10.0
15 : 8.0
16 : 6.000000000000001
17 : 3.999999999999999
18 : 1.9999999999999996
19 : 0.0
</code></pre>