如何使用{value:(row#,column#)}对字典替换2D numpy数组中的值 [英] How do I replace values in 2D numpy array using a dictionary of {value:(row#,column#)} pairs
问题描述
import numpy as np
数组看起来像这样:
array = np.zeros((10,10))
array =
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
字典是这样的:
dict = {72: (3, 4), 11: (1, 5), 10: (2, 4), 43: (2, 3), 22: (24,35), 11: (8, 9)}
我要遍历数组,并将与字典中的网格坐标匹配的所有栅格点替换为字典中的相应值
I want to iterate over the array and replace any grid points that match the grid coordinates in the dictionary with the corresponding value in the dictionary
我在这样的输出之后:
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 11. 0. 0. 0. 0.]
[ 0. 0. 0. 43. 10. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 72. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 11.]
[ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]]
** 我已经编辑了问题,提供位于阵列内的坐标,除了 1 个例外.我还提供了所需输出的示例
** i have edited the question the provide co-ordinates that sit within the array except for 1 exception. I also provided an example of the desired output
推荐答案
在替换数组之前,我们需要先将值=> coordinates转换为coordinates => values的映射.我已经编辑了字典条目以进行演示,并且正如注释中所指出的,字典坐标条目应小于数组的尺寸
We need to invert mapping from values=>coordinates to co-ordinates=>values before replacement in the array. I have edited the dictionary entries for demo purpose and as pointed in comments, dictionary co-ordinate entries should be less than dimensions of array
import numpy as np
arrObj = np.zeros((10,10))
arrObj
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
# [ 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
#copy of array for replacement
replaceArrObj=arrObj
#ensure co-ordinates in the dictionary could be indexed in array
#current mapping: values => co-ordinates
dictObj = {1.0:(0.0,0.0),2.0:(1.0,1.0),3.0: (2.0, 2.0), 4.0: (3.0, 3.0),5.0:(4.0,4.0), 6.0: (5.0, 5.0), 7.0: (6.0, 6.0), 8.0: (7.0,7.0), 9.0: (8.0,8.0),
10.0: (9.0,9.0)}
dictObj
#{1.0: (0.0, 0.0),
# 2.0: (1.0, 1.0),
# 3.0: (2.0, 2.0),
# 4.0: (3.0, 3.0),
# 5.0: (4.0, 4.0),
# 6.0: (5.0, 5.0),
# 7.0: (6.0, 6.0),
# 8.0: (7.0, 7.0),
# 9.0: (8.0, 8.0),
# 10.0: (9.0, 9.0)}
反转映射:
#invert mapping of dictionary: co-ordinates => values
inv_dictObj = {v: k for k, v in dictObj.items()}
inv_dictObj
#{(0.0, 0.0): 1.0,
# (1.0, 1.0): 2.0,
# (2.0, 2.0): 3.0,
# (3.0, 3.0): 4.0,
# (4.0, 4.0): 5.0,
# (5.0, 5.0): 6.0,
# (6.0, 6.0): 7.0,
# (7.0, 7.0): 8.0,
# (8.0, 8.0): 9.0,
# (9.0, 9.0): 10.0}
替换:
#Replace values from dictionary at correponding co-ordiantes
for i,j in inv_dictObj.keys():
replaceArrObj[i,j]=inv_dictObj[(i,j)]
replaceArrObj
#array([[ 1., 0., 0., 0., 0., 0., 0., 0., 0., 0.],
# [ 0., 2., 0., 0., 0., 0., 0., 0., 0., 0.],
# [ 0., 0., 3., 0., 0., 0., 0., 0., 0., 0.],
# [ 0., 0., 0., 4., 0., 0., 0., 0., 0., 0.],
# [ 0., 0., 0., 0., 5., 0., 0., 0., 0., 0.],
# [ 0., 0., 0., 0., 0., 6., 0., 0., 0., 0.],
# [ 0., 0., 0., 0., 0., 0., 7., 0., 0., 0.],
# [ 0., 0., 0., 0., 0., 0., 0., 8., 0., 0.],
# [ 0., 0., 0., 0., 0., 0., 0., 0., 9., 0.],
# [ 0., 0., 0., 0., 0., 0., 0., 0., 0., 10.]])
类型转换:
只要数组坐标和字典条目的类型相同,您就不会遇到任何错误/警告.如果您更喜欢int/float
You should not face any errors/warnings as long as both array co-ordinates and dictionary entries have same type. You can additionally enforce specific type conversion if you prefer int/float
#float to int conversion in array
replaceArrObj.astype(int)
#array([[ 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
# [ 0, 2, 0, 0, 0, 0, 0, 0, 0, 0],
# [ 0, 0, 3, 0, 0, 0, 0, 0, 0, 0],
# [ 0, 0, 0, 4, 0, 0, 0, 0, 0, 0],
# [ 0, 0, 0, 0, 5, 0, 0, 0, 0, 0],
# [ 0, 0, 0, 0, 0, 6, 0, 0, 0, 0],
# [ 0, 0, 0, 0, 0, 0, 7, 0, 0, 0],
# [ 0, 0, 0, 0, 0, 0, 0, 8, 0, 0],
# [ 0, 0, 0, 0, 0, 0, 0, 0, 9, 0],
# [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 10]])
#float to int conversion in dictionary, where k referes to key items and v to value items
int_dictObj = { (int(k[0]),int(k[1])):int(v) for k,v in inv_dictObj.items()}
int_dictObj
#{(0, 0): 1,
# (1, 1): 2,
# (2, 2): 3,
# (3, 3): 4,
# (4, 4): 5,
# (5, 5): 6,
# (6, 6): 7,
# (7, 7): 8,
# (8, 8): 9,
# (9, 9): 10}
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