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)}
我想迭代数组，并将与字典中的网格坐标匹配的任何网格点替换为字典中的对应值
我正在追求这样的输出：
[ 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.]]
**我编辑了问题，除了1个例外，提供坐在数组中的坐标。我还提供了所需输出的示例
答案 0 :(得分：3)
我希望我能正确理解你的问题
array = np.zeros((10,10))
data = {72: (3, 4), 11: (1, 5), 10: (2, 4), 43: (2, 3), 22: (24,35)}
for i in data.keys():
try:
array[data[i][0],data[i][1]] = float(i)
except IndexError:
pass
print array
我更改了索引，使其适合您的10 x 10阵列（我假设您在实际示例中使用了更大的数组）
我遍历字典中的所有键（值）。然后程序尝试在给定坐标的数组中设置此值。 我为一些坐标在数组外部的情况传递了IndexErrors（就像这个例子中的最后一个。
修改
此解决方案仅在您的密钥是唯一的情况下才有效。如果不是，我会推荐@Osssan的解决方案。
答案 1 :(得分：1)
我们需要在数组中替换之前将值从值=＆gt;坐标转换为坐标=＆gt;值。我已经编辑了字典条目以用于演示目的，并且如评论中所指出的，字典坐标条目应该小于数组的维度
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}
<强>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具有相同类型，就不应该遇到任何错误/警告。 如果您更喜欢int / float
，还可以强制执行特定的类型转换#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}