条件numpy.cumsum？

Question

我对python和numpy非常陌生，如果我误用了一些术语，我很抱歉。

我已经将光栅转换成二维数字阵列，希望能快速高效地对其进行计算。

• 我需要获得一个numpy数组的累积和，以便为每个值生成小于或等于该值的所有值的和，并将该值写入一个新数组。我需要这样循环整个数组。
• 我还需要在1到100之间扩展输出，但这似乎是 直截了当。

试图通过实例澄清：

array([[ 4,  1  ,  3   ,  2] dtype=float32)

我希望输出值(只需手工进行第一行)读到：

array([[ 10,  1  ,  6   ,  3], etc.

对怎么做有什么想法吗？

提前感谢！

对于任何感兴趣的人来说，即将完成的脚本：

#Generate Cumulative Thresholds
#5/15/14

import os
import sys
import arcpy
import numpy as np

#Enable overwriting output data
arcpy.env.overwriteOutput=True

#Set working directory
os.chdir("E:/NSF Project/Salamander_Data/Continuous_Rasters/Canadian_GCM/2020/A2A/")

#Set geoprocessing variables
inRaster = "zero_eurycea_cirrigera_CA2A2020.tif"
des = arcpy.Describe(inRaster)
sr = des.SpatialReference
ext = des.Extent
ll = arcpy.Point(ext.XMin,ext.YMin)

#Convert GeoTIFF to numpy array
a = arcpy.RasterToNumPyArray(inRaster)

#Flatten for calculations
a.flatten()

#Find unique values, and record their indices to a separate object
a_unq, a_inv = np.unique(a, return_inverse=True)

#Count occurences of array indices
a_cnt = np.bincount(a_inv)

#Cumulatively sum the unique values multiplied by the number of
#occurences, arrange sums as initial array
b = np.cumsum(a_unq * a_cnt)[a_inv]

#Divide all values by 10 (reverses earlier multiplication done to
#facilitate accurate translation of ASCII scientific notation
#values < 1 to array)
b /= 10

#Rescale values between 1 and 100
maxval = np.amax(b)
b /= maxval
b *= 100

#Restore flattened array to shape of initial array
c = b.reshape(a.shape)

#Convert the array back to raster format
outRaster = arcpy.NumPyArrayToRaster(c,ll,des.meanCellWidth,des.meanCellHeight)

#Set output projection to match input
arcpy.DefineProjection_management(outRaster, sr)

#Save the raster as a TIFF
outRaster.save("C:/Users/mkcarte2/Desktop/TestData/outRaster.tif")

sys.exit()

Answer 1

根据您想要处理重复的方式，这可能会起作用：

In [40]: a
Out[40]: array([4, 4, 2, 1, 0, 3, 3, 1, 0, 2])

In [41]: a_unq, a_inv = np.unique(a, return_inverse=True)

In [42]: a_cnt = np.bincount(a_inv)

In [44]: np.cumsum(a_unq * a_cnt)[a_inv]
Out[44]: array([20, 20,  6,  2,  0, 12, 12,  2,  0,  6], dtype=int64)

当然，a是您的数组扁平化的地方，然后您必须将其重塑为原始形状。

当然，一旦numpy 1.9退出，您就可以将上面的41和42线压缩成一个更快的单行：

a_unq, a_inv, a_cnt = np.unique(a, return_inverse=True, return_counts=True)

Answer 2

编辑：

这很难看，但我认为它终于奏效了：

import numpy as np

def cond_cum_sum(my_array):
    my_list = []
    prev = -np.inf
    prev_sum = 0
    for ele in my_array:
        if prev != ele:
            prev_sum += ele
        my_list.append(prev_sum)
        prev = ele
    return np.array(my_list)

a = np.array([[4,2,2,3],
              [9,0,5,2]], dtype=np.float32)

flat_a = a.flatten()
flat_a.sort() 

temp = np.argsort(a.ravel())   

cum_sums = cond_cum_sum(flat_a)

result_1 = np.zeros(len(flat_a))
result_1[temp] = cum_sums

result = result_1.reshape(a.shape)

结果：

>>> result
array([[  9.,   2.,   2.,   5.],
       [ 23.,   0.,  14.,   2.]])

Answer 3

使用更少的矮小和更多的python：

a = np.array([[4,2,2,3],
              [9,0,5,2]], dtype=np.float32)

np.array([[sum(x for x in arr if x <= subarr) for subarr in arr] for arr in a])
# array([[ 11.,   4.,   4.,   7.],
#        [ 16.,   0.,   7.,   2.]])

如果和只考虑项目一次，无论项目出现多少，那么，

np.array([[sum(set(x for x in arr if x <= subarr)) for subarr in arr] for arr in a]) 
# array([[  9.,   2.,   2.,   5.],
#        [ 16.,   0.,   7.,   2.]])