平均数据箱中的数据

时间:2018-04-15 16:54:55

标签: python python-3.x numpy average scientific-computing

我有两个列表:1是深度列表,另一个是叶绿素列表,它们彼此对应。我希望每0.5米深度平均一次叶绿素数据。

chl  = [0.4,0.1,0.04,0.05,0.4,0.2,0.6,0.09,0.23,0.43,0.65,0.22,0.12,0.2,0.33]
depth = [0.1,0.3,0.31,0.44,0.49,1.1,1.145,1.33,1.49,1.53,1.67,1.79,1.87,2.1,2.3]

深度箱的长度并不总是相等,并且不总是以0.0或0.5的间隔开始。叶绿素数据总是与深度数据协调。叶绿素平均值也不能按升序排列,需要根据深度保持正确的顺序。深度和叶绿素列表很长,所以我不能单独这样做。

我如何制作0.5米深的垃圾箱,其中包含平均叶绿素数据?

目标:

depth = [0.5,1.0,1.5,2.0,2.5]
chlorophyll = [avg1,avg2,avg3,avg4,avg5]

例如:

avg1 = np.mean(0.4,0.1,0.04,0.05,0.4)

4 个答案:

答案 0 :(得分:4)

这是一个向量化的NumPy解决方案,使用np.searchsorted获取bin移位(索引)和np.add.reduceat用于分箱摘要 -

def bin_data(chl, depth, bin_start=0, bin_length= 0.5):
    # Get number of intervals and hence the bin-length-spaced depth array
    n = int(np.ceil(depth[-1]/bin_length))
    depthl = np.linspace(start=bin_start,stop=bin_length*n, num=n+1)

    # Indices along depth array where the intervaled array would have bin shifts
    idx = np.searchsorted(depth, depthl)

    # Number of elements in each bin (bin-lengths)
    lens = np.diff(idx)

    # Get summations for each bins & divide by bin lengths for binned avg o/p
    # For bins with lengths==0, set them as some invalid specifier, say NaN
    return np.where(lens==0, np.nan, np.add.reduceat(chl, idx[:-1])/lens)

示例运行 -

In [83]: chl
Out[83]: 
array([0.4 , 0.1 , 0.04, 0.05, 0.4 , 0.2 , 0.6 , 0.09, 0.23, 0.43, 0.65,
       0.22, 0.12, 0.2 , 0.33])

In [84]: depth
Out[84]: 
array([0.1  , 0.3  , 0.31 , 0.44 , 0.49 , 1.1  , 1.145, 1.33 , 1.49 ,
       1.53 , 1.67 , 1.79 , 1.87 , 2.1  , 2.3  ])

In [85]: bin_data(chl, depth, bin_start=0, bin_length= 0.5)
Out[85]: array([0.198,   nan, 0.28 , 0.355, 0.265])

答案 1 :(得分:4)

我很惊讶scipy.stats.binned_statistic尚未被提及。您可以直接用它计算平均值,并使用可选参数指定容器。

from scipy.stats import binned_statistic

mean_stat = binned_statistic(depth, chl, 
                             statistic='mean', 
                             bins=5, 
                             range=(0, 2.5))

mean_stat.statistic
# array([0.198,   nan, 0.28 , 0.355, 0.265])
mean_stat.bin_edges
# array([0. , 0.5, 1. , 1.5, 2. , 2.5])
mean_stat.binnumber
# array([1, 1, 1, ..., 4, 5, 5])

答案 2 :(得分:3)

一种方法是使用numpy.digitize来分类您的类别。

然后使用字典或列表推导来计算结果。

import numpy as np

chl  = np.array([0.4,0.1,0.04,0.05,0.4,0.2,0.6,0.09,0.23,0.43,0.65,0.22,0.12,0.2,0.33])
depth = np.array([0.1,0.3,0.31,0.44,0.49,1.1,1.145,1.33,1.49,1.53,1.67,1.79,1.87,2.1,2.3])

bins = np.array([0,0.5,1.0,1.5,2.0,2.5])

A = np.vstack((np.digitize(depth, bins), chl)).T

res = {bins[int(i)]: np.mean(A[A[:, 0] == i, 1]) for i in np.unique(A[:, 0])}

# {0.5: 0.198, 1.5: 0.28, 2.0: 0.355, 2.5: 0.265}

或者您所使用的精确格式:

res_lst = [np.mean(A[A[:, 0] == i, 1]) for i in range(len(bins))]

# [nan, 0.198, nan, 0.28, 0.355, 0.265]

答案 3 :(得分:3)

以下是I/flutter (26289): null

的一种方法
pandas.cut