带条件求和的 Python 代码
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Python code for sum with condition
提问by beistvan
The task is following: sum the list elements with even indexes and multiply the result by the last list's elemet. I have this oneliner solution code in Python.
任务如下：对具有偶数索引的列表元素求和，并将结果乘以最后一个列表的元素。我在 Python 中有这个 oneliner 解决方案代码。
array = [-37,-36,-19,-99,29,20,3,-7,-64,84,36,62,26,-76,55,-24,84,49,-65,41]
print sum(i for i in array if array.index(i) % 2 == 0)*array[-1] if array != [] else 0
My result is -1476 ( The calculation is: 41*(-37-19+29+3-64+36+26+55+84-65) )
我的结果是-1476（计算是：41*(-37-19+29+3-64+36+26+55+84-65)）
The right result is 1968.
正确的结果是 1968 年。
I can't figure it out why this code is not working correctly in this particular case.
我无法弄清楚为什么此代码在这种特殊情况下无法正常工作。
采纳答案by YS-L
There is a repeated element 84in the list, thus array.indexdoes not work as you expected it to be. Also, your code has a quadratic complexity which is not required.
列表中有一个重复的元素84，因此array.index不会像您预期的那样工作。此外，您的代码具有不需要的二次复杂度。
To fix your code with a minimum amount of edit, it would look something like this:
要以最少的编辑量修复您的代码，它看起来像这样：
array = [-37,-36,-19,-99,29,20,3,-7,-64,84,36,62,26,-76,55,-24,84,49,-65,41]
print sum(array[i] for i in range(len(array)) if i % 2 == 0)*array[-1] if array != [] else 0
回答by shaktimaan
This is what you are looking for:
这就是你要找的：
array[-1] * sum(array[::2])
array[::2]traverses the array from first index to last index in steps of two, i.e., every alternate number. sum(array[::2])gets the sum of alternate numbers from the original list.
array[::2]从第一个索引到最后一个索引以两步（即每个备用数）遍历数组。sum(array[::2])从原始列表中获取备用数字的总和。
Using indexwill work as expected only when you are sure the list does not have duplicates, which is why your code fails to give the correct result.
index只有当您确定列表没有重复项时，使用才会按预期工作，这就是您的代码无法给出正确结果的原因。
回答by vaultah
>>> sum(x for i, x in enumerate(array) if not i % 2)*array[-1]
1968
Use the built-in enumeratefunction, since there're duplicate elements in your list, and list.index(x)returns the index of the firstelement equal to x(as said in the documentation). Also take a look at the documentation on enumerate.
使用内置enumerate函数，因为列表中有重复元素，并list.index(x)返回第一个元素的索引等于x（如文档中所述）。另请查看有关enumerate.
