Question

如何从其他列表创建新的运行总计列表？假设我有以下问题：

true
true
true
true
false
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false

我的数据采用这种格式（sumbl3）：

sumbl4 = LaborDetail.objects.filter(businessunit='Birmingham').filter(endofweek__gte=datetime.date.today()-datetime.timedelta(days=60)).values_list('endofweek').annotate(hours__sum=Sum('hours')).order_by('endofweek')

sumbl3 = json.dumps(list(sumbl4), cls=DecimalJSONEncoder)

所以我的新列表看起来像是：

[["2017/04/23", 972.5], ["2017/04/30", 1076.5], ["2017/05/07", 1162.5], ["2017/05/14", 1055.5], ["2017/05/21", 981.0], ["2017/05/28", 945.5], ["2017/06/04", 912.0], ["2017/06/11", 1106.0], ["2017/06/18", 1059.0]]

注意：我使用的是Python 2.7

Answer 1

这个怎么样：

orig = [["2017/04/23", 972.5], ["2017/04/30", 1076.5], ["2017/05/07", 1162.5], ["2017/05/14", 1055.5], ["2017/05/21", 981.0], ["2017/05/28", 945.5], ["2017/06/04", 912.0], ["2017/06/11", 1106.0], ["2017/06/18", 1059.0]]

new = []
temp_sum = 0
for o in orig:
  a = o[0]
  b = o[1] + temp_sum
  new.append([a, b])
  temp_sum += b

print new

结果：

[['2017/04/23', 972.5], ['2017/04/30', 2049.0], ['2017/05/07', 4184.0], ['2017/05/14', 8261.0], ['2017/05/21', 16447.5], ['2017/05/28', 32859.5], ['2017/06/04', 65685.5], ['2017/06/11', 131565.0], ['2017/06/18', 263083.0]]

Answer 2

另一种让你获得最终名单的方法：

a = [['2017/04/23', 972.5],
 ['2017/04/30', 1076.5],
 ['2017/05/07', 1162.5],
 ['2017/05/14', 1055.5],
 ['2017/05/21', 981.0],
 ['2017/05/28', 945.5],
 ['2017/06/04', 912.0],
 ['2017/06/11', 1106.0],
 ['2017/06/18', 1059.0]]


sub = [a[0]]

for k, v in a[1:]:
    b = sub[-1][1] + v
    sub.append([k, b])

print(sub)

>>> [['2017/04/23', 972.5],
 ['2017/04/30', 2049.0],
 ['2017/05/07', 3211.5],
 ['2017/05/14', 4267.0],
 ['2017/05/21', 5248.0],
 ['2017/05/28', 6193.5],
 ['2017/06/04', 7105.5],
 ['2017/06/11', 8211.5],
 ['2017/06/18', 9270.5]]

或者以更紧凑的方式，您可以创建一个方法/函数，如下例所示：

def get_sum(a):
    sub = []
    for k, v in a:
        if not sub:
            sub.append([k, v])
        else:
             sub.append([k, sub[-1][1] +v])
    return sub

Answer 3

虽然Python 2.7中不包含itertools.accumulate，但您可以使用文档中提供的配方。

import operator

def accumulate(iterable, func=operator.add):
    'Return running totals'
    # accumulate([1,2,3,4,5]) --> 1 3 6 10 15
    # accumulate([1,2,3,4,5], operator.mul) --> 1 2 6 24 120
    it = iter(iterable)
    try:
        total = next(it)
    except StopIteration:
        return
    yield total
    for element in it:
        total = func(total, element)
        yield total

然后，它的用法如下：

data = [(1, 1), (2, 2), (3, 3)]

keys, values = zip(*data)
accumulated_data = zip(keys, accumulate(values))
# [(1, 1), (2, 3), (3, 6)]

Answer 4

这个解决方案给了我正在寻找的东西：

sumbl4 = LaborDetail.objects.filter(businessunit='Birmingham').filter(endofweek__gte=datetime.date.today()-datetime.timedelta(days=60)).values_list('endofweek').annotate(hours__sum=Sum('hours')).order_by('endofweek')

sumbl3 = json.dumps(list(sumbl4), cls=DecimalJSONEncoder)

list_of_running_totals = []
running_total = 0
for date, amount in sumbl4:
    running_total += amount
    list_of_running_totals.append([date, running_total])
print list_of_running_totals

Django - 从现有列表中创建一个新的运行总计列表

4 个答案: