我每小时都有关于自行车租赁需求和天气的数据。我想分别绘制每个小时的平均需求，并分别计算好天气和坏天气。
当我绘制给定时间的平均需求（不考虑天气）时，我所做的是计算给定时间的租金总需求，然后除以总小时数：
hour_count = np.bincount(hour)
for i in range(number_of_observations):
hour_sums[hour[i]] = hour_sums[hour[i]] + rentals[i]
av_rentals = [x/y for x,y in zip(hour_sums,hour_count)]
现在，我想这样做，但要分别考虑好天气和坏天气。累计和很容易，我刚刚添加了一个“ if”子句。我不知道该如何计算好坏天气的小时数。我宁愿避免像sum那样做大循环……任何与bincount相同但带有子句的函数？就像是：
good_weather_hour_count = np.bincount(hour, weather == 1 or weather == 2)
有任何想法吗？
PS。也许有人知道如何在给定的小时数内不加循环地汇总租金？我尝试使用2d直方图进行操作，但是没有用。
label_sums = np.histogram2d(hour, rentals, bins=24)[0]
np.bincount有一个weights参数，您可以使用该参数对以租借数量加权的小时数进行二进制计数。例如，
In [39]: np.bincount([1,2,3,1], weights=[20,10,40,10])
Out[39]: array([ 0., 30., 10., 40.])
因此，您可以替换for-loop：
for i in range(number_of_observations):
hour_sums[hour[i]] = hour_sums[hour[i]] + rentals[i]
和
hour_sums = np.bincount(hour, weights=rentals, minlength=24)
要处理好/坏天气，您可以屏蔽hour和rentals数据，以仅选择适用的数据子集：
mask = (weather == w)
masked_hour = hour[mask]
masked_rentals = rentals[mask]
然后在masked_hour和上进行计算masked_rentals：
import numpy as np
np.random.seed(2016)
N = 2
hour = np.tile(np.arange(24), N)
rentals = np.random.randint(10, size=(len(hour),))
# say, weather=1 means good weather, 2 means bad weather
weather = np.random.randint(1, 3, size=(len(hour),))
average_rentals = dict()
for kind, w in zip(['good', 'bad', 'all'], [1, 2, None]):
if w is None:
mask = slice(None)
else:
mask = (weather == w)
masked_hour = hour[mask]
masked_rentals = rentals[mask]
total_rentals = np.bincount(masked_hour, weights=masked_rentals, minlength=24)
total_hours = np.bincount(masked_hour, minlength=24)
average_rentals[kind] = (total_rentals / total_hours)
for kind, result in average_rentals.items():
print('\n{}: {}'.format(kind, result))
产量
bad: [ 4. 6. 2. 5.5 nan 4. 4. 8. nan 3. nan 2.5 4. nan 9.
nan 3. 5.5 8. nan 8. 5. 9. 4. ]
good: [ 3. nan 4. nan 8. 4. nan 7. 5.5 2. 4. nan nan 0.5 9.
0.5 nan nan 5. 7. 1. 7. 8. 0. ]
all: [ 3.5 6. 3. 5.5 8. 4. 4. 7.5 5.5 2.5 4. 2.5 4. 0.5 9.
0.5 3. 5.5 6.5 7. 4.5 6. 8.5 2. ]
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