Question

我正在尝试获取列表清单的总和，其中下面定义的列表的输出为[6,2,10]。

SELECT  e.id` AS id, e.first_name AS first_name,
        u.initials AS assigned_coach,
        e.gender AS gender, e.email AS email, e.phone AS phone,
        e.age AS age, e.version AS version,
        e.evaluation_status AS evaluation_status,
        e.ip_address AS ip_address, e.date_created AS date_created,
        e.date_updated AS date_updated
    FROM  evaluation_client AS e
    LEFT JOIN  cx_user AS u  ON e.cx_hc_user_id = u.id
    ORDER BY  e.id ASC
    LIMIT  :APL0 OFFSET, :APL1

但是，此程序的结果为[6，8，18]。任何帮助将不胜感激。

Answer 1

使用list comprehension。

列表中的每个元素都是iterable，因此，sum是完成此任务的最佳选择。

>>> data = [[1,2,3], [2], [1, 2, 3, 4]]
>>> [sum(d) for d in data]
[6, 2, 10]

现在，如果您想知道问题出在哪里...

在 data 循环后放置 accumulator ，并在每个 row 循环后填充列表：

>>> data = [[1,2,3], [2], [1, 2, 3, 4]]
>>> output = []
>>> for row in data:
...     total = 0
...     for val in row[0:len(row)]:
...         total += val
...     output.append(total)
...
>>> output
[6, 2, 10]

Answer 2

使用map()：

data = [[1,2,3], [2], [1, 2, 3, 4]] 
print(list(map(sum, data)))
# [6, 2, 10]

或列表理解：

data = [[1,2,3], [2], [1, 2, 3, 4]]
print([sum(x) for x in data])
# [6, 2, 10]

如何打印包含单独的列表总和的列表

2 个答案: