如何计算分位数中的观测数?

时间:2019-01-31 00:10:43

标签: r statistics probability frequency quantile

考虑一下,根据参数(3,5)的Gamma分布,我有100万个观测值。我可以使用summary()来找到分位数,但是我试图找出每个红线之间有多少个观测值,这些观测值被分为10条?

a = rgamma(1e6, shape = 3, rate = 5)

summary(a)

  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
0.0053  0.3455  0.5351  0.6002  0.7845  4.4458

enter image description here

1 个答案:

答案 0 :(得分:3)

我们可以将cuttable一起使用:

table(cut(a, quantile(a, 0:10 / 10)))

# (0.00202,0.22]   (0.22,0.307]  (0.307,0.382]  (0.382,0.457]  (0.457,0.535]  (0.535,0.622] 
#          99999         100000         100000         100000         100000         100000 
#  (0.622,0.724]  (0.724,0.856]   (0.856,1.07]    (1.07,3.81] 
#         100000         100000         100000         100000 

但是考虑到分位数是多少,这可能不是很有趣。也许您可能也想尝试理论分位数:

table(cut(a, qgamma(0:10 / 10, 3, 5)))

#      (0,0.22]  (0.22,0.307] (0.307,0.383] (0.383,0.457] (0.457,0.535] (0.535,0.621] (0.621,0.723] 
#         99978        100114        100545         99843         99273         99644        100104 
# (0.723,0.856]  (0.856,1.06]    (1.06,Inf] 
#        100208         99883        100408 

没什么有趣的,因为,如果您的数据确实遵循伽马分布并且您有大量观察结果,那么您可以确定在第q次与(q + x)-理论分位数。在较小的样本中,第二种方法可能很有趣。


编辑:针对您的最新问题,很明显,有10%,20%的意思不是分位数。假设最小值为0,最大值为2,如果占10%,您认为0.2,那么您想要

table(cut(a, seq(min(a), max(a), length = 10 + 1)))

# (0.00418,0.428]   (0.428,0.853]    (0.853,1.28]      (1.28,1.7]      (1.7,2.13]     (2.13,2.55] 
#          361734          436176          155332           37489            7651            1335 
#     (2.55,2.97]      (2.97,3.4]      (3.4,3.82]     (3.82,4.25] 
#             231              38              11               2