Question

我对python和numpy很新，很抱歉如果我误用了一些术语。

我已将光栅转换为2D numpy数组，希望能快速高效地对其进行计算。

我需要得到一个numpy数组的累积和，以便每个值，我生成小于或的所有值的总和等于，并将该值写入新数组。我需要循环通过整个阵列。
我还需要在1到100之间缩放输出，但这似乎是更直截了当。

尝试通过示例澄清：

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

我希望输出值（只是手动执行第一行）：

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

关于如何做到这一点的任何想法？

提前致谢！

对于任何感兴趣的人来说，已完成的剧本近在咫尺：

#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

根据您希望如何处理重复，这可能有效：

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阵列被展平，你必须重新塑造成原始形状。

当然，一旦numpy 1.9出局，你可以将上面的第41和42行压缩成单一的，更快的：

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

Answer 2

修改

这很难看，但我认为它最终有效：

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

更少的numpy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.]])

如果总和只考虑项目一次，无论它们出现多少，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.]])

Answer 4

这个怎么样：

a=np.array([ 4,  1  ,  3   ,  2])

np.array([np.sum(a[a<=x])for x in a])

给出

array([10,  1,  6,  3])

对于2D数组（假设您想要整个数组的总和，而不仅仅是行）：

a=np.array([[ 4,  1  ,  3   ,  2],[ 5,  1  ,  3   ,  2]])

np.array([[np.sum(a[a<=x])for x in a[y,:]]for y in range(a.shape[0])])

Gvies

array([[16,  2, 12,  6],
       [21,  2, 12,  6]])

有条件的numpy.cumsum？

4 个答案: