我怎樣才能得到結果和abResult?請引導我通過,我真的很感激它!基於索引列表求和列表元素
x=[1,2]
y=[3,4,5]
result= [3,9] <=== the sum of result determine by x
============= OR ==========
a=[1,3,2]
b= [4,2,3,4,5,10]
abResult= [4,9,15]
我怎樣才能得到結果和abResult?請引導我通過,我真的很感激它!基於索引列表求和列表元素
x=[1,2]
y=[3,4,5]
result= [3,9] <=== the sum of result determine by x
============= OR ==========
a=[1,3,2]
b= [4,2,3,4,5,10]
abResult= [4,9,15]
a=[1,3,2]
b= [4,2,3,4,5,10]
res=[]
for i in a:
res.append(sum(b[:i]))
b=b[i:]
資源存儲您的結果
是的,它的確有用。謝謝!! – Kily
這是可能工作正常。
def isValid(counter, array): # valid if sum of counter equals array length
return sum(counter) == len(array)
def calc(counter, array):
if not isValid(counter, array):
return []
flag = 0
for x in counter:
yield sum(array[flag:flag+x]) # sum between (flag ~ flag+x)
flag += x # add x to flag for next flag
我們應該知道如何從前兩個列表中獲得結果嗎?或者只是猜測? –
所以基本上,a [0] = 1然後你總結(b [1])或a [1] = 3然後你總結(2,3,4)b,希望它是有道理的。 – Kily