Question

我每小时有关自行车租赁和天气需求的数据。我想描绘每小时的平均需求，分别是好天气和恶劣天气。

当我在给定时间绘制平均需求（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)]

现在我想做同样的事情，但是分开来看好天气和恶劣天气。累积总和很简单，我刚刚添加了一个'if'子句。我不知道如何计算好几天和坏天气。我宁愿避免做一个像总和一样的大循环...任何与bincount相同但带有子句的函数？类似的东西：

good_weather_hour_count = np.bincount(hour, weather == 1 or weather == 2)

任何想法？
PS。也许有人知道如何在没有循环的情况下对给定小时的租金进行总结？我用2d直方图尝试了一些东西，但它没有用。

label_sums = np.histogram2d(hour, rentals, bins=24)[0]

Answer 1

np.bincount has a weights parameter您可以使用它来计算按租金数量加权的小时数。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]

带

hour_sums = np.bincount(hour, weights=rentals, minlength=24)

为了处理好/坏天气，您可以屏蔽hour和rentals数据，以便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masked_rentals上进行计算：

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. ]

Answer 2

我不确定Numpy，但你可以用标准库轻松地做到这一点：

from collections import Counter, defaultdict

weather_counts = defaultdict(Counter)

times = [
    {'time': '1:00 AM', 'weather': 1},
    {'time': '2:00 AM', 'weather': 2},
    {'time': '5:00 PM', 'weather': 2},
    {'time': '3:00 AM', 'weather': 1},
    {'time': '1:00 AM', 'weather': 1},
]

rentals = [
    1,
    2,
    5,
    3,
    3,
]

for times, rental_count in zip(times, rentals):
    weather_counts[times['weather']][times['time']] += rental_count

import pprint; pprint.pprint(weather_counts)

有没有办法在python中使用带有子句的bincount？

2 个答案: