目标是将一维数组视为二维网格。第二个一维数组给出了需要在网格中更改的值列表，第三个数组指示了多少。
要注意的是修改值周围的值也会更改。
下面的示例仍然是一维数组，但对其进行计算就好像它是二维网格一样。有用;但目前它会更改网格中与一维列表(示例)中的值匹配的所有值。对于列表中的 1 个值，我不想只转换 1 个值及其周围环境。
即如果列表是[2,3]；我只想更改迭代中遇到的第一个 2 和 3 值。目前的例子，改变网格中的每 2 个。
让我感到困惑的是(可能是因为我构造修改计算的方式)，我不能简单地遍历网格并在每次匹配时删除列表值。
预先感谢您的宝贵时间!
代码如下；
import numpy
def grid_range(value):
if value > 60000:
value = 60000
return (value)
elif value < 100:
value = 100
return(value)
elif value <= 60000 and value >= 100:
return(value)
def grid(array,samples,details):
original_length = len(array)
c = int((original_length)**0.5)
new_array = [] #create a new array with the modified values
for elem in range (len(array)): #if the value is in samples
if array[elem] in samples:
value = array[elem] + (array[elem] * (details[1]/100))
test_range = grid_range(value)
new_array.append(test_range)
elif ((elem + 1) < original_length) and array[elem - 1] in samples: #change the one before the value
if (len(new_array) % c == 0) and array[elem + 1] not in samples:
new_array.append(array[elem])
else:
new_forward_element = array[elem] +(array[elem] * (details[2]/100))
test_range1 = grid_range(new_forward_element)
new_array.append(test_range1)
elif ((elem + 1) < original_length) and (array[elem + 1]) in samples: #change the one before and that it doesn't attempt to modify passed the end of the array
if (len(new_array) + 1) % c == 0:
new_array.append(array[elem])
else:
new_back_element = array[elem] +(array[elem] * (details[2]/100))
test_range2 = grid_range(new_back_element)
new_array.append(test_range2)
elif ((elem+c) <= (original_length - c))and(array[elem + c]) in samples: #if based on the 9 numbers on the right of the keyboard with test value numebr 5; this is position '2'
extra1 = array[elem] +(array[elem] * (details[2]/100))
test_range3 = grid_range(extra1)
new_array.append(test_range3)
elif (array[abs(elem - c)]) in samples: #position '8'
extra2 = array[elem] +(array[elem] * (details[2]/100))
test_range4 = grid_range(extra2)
new_array.append(test_range4)
elif (array[abs(elem - (c-1))]) in samples: #position '7'
if (elem - (c-1)) % c == 0:
new_array.append(array[elem])
else:
extra3 = array[elem] +(array[elem] * (details[2]/100))
test_range5 = grid_range(extra3)
new_array.append(test_range5)
elif (array[abs(elem - (c+1))]) in samples: #position '9'
if (elem - (c+1) + 1) % c == 0:
new_array.append(array[elem])
else:
extra4 = array[elem] +(array[elem] * (details[2]/100))
test_range6 = grid_range(extra4)
new_array.append(test_range6)
elif ((elem +(c-1)) < original_length) and (array[elem + (c-1)]) in samples: #position '1', also not passed total array length
if (elem + (c-1)+ 1) % c == 0:
new_array.append(array[elem])
else:
extra5 = array[elem] +(array[elem] * (details[2]/100))
test_range7 = grid_range(extra5)
new_array.append(test_range7)
elif (elem + (c+1)) < (len(array)- c) and (array[elem + (c+1)]) in samples: #position '3', also not passed total array length
if (elem + (c+1)) % c == 0:
new_array.append(array[elem])
else:
extra6 = array[elem] +(array[elem] * (details[2]/100))
test_range8 = grid_range(extra6)
new_array.append(test_range8)
else:
new_array.append(array[elem])
return(new_array)
a = [16,2,20,4,14,6,70,8,9,100,32,15,7,14,50,20,17,10,9,20,7,17,50,2,19,20]
samples = [2]
grid_details = [10,50,100]
result = grid(a,samples,grid_details)
编辑:
根据你的回答 Joe，我创建了一个版本，它通过特定的 % 修改主要值(中心)，并通过另一个修改周围的元素。但是，如何确保在下一次样本迭代期间不会再次转换更改的值。
感谢您的宝贵时间!
示例代码:
def grid(array,samples,details):
#Sides of the square (will be using a squarable number
Width = (len(array)) ** 0.5
#Convert to grid
Converted = array.reshape(Width,Width)
#Conversion details
Change = [details[1]] + [details[2]]
nrows, ncols = Converted.shape
for value in samples:
#First instance where indexing returns it
i,j = np.argwhere(Converted == value)[0]
#Prevent indexing outside the boudaries of the
#array which would cause a "wraparound" assignment
istart, istop = max(i-1, 0), min(i+2, nrows)
jstart, jstop = max(j-1, 0), min(j+2, ncols)
#Set the value within a 3x3 window to their "new_value"
for elem in Converted[istart:istop, jstart:jstop]:
Converted[elem] = elem + (elem * (value * ((Change[1]/100))
#Set the main value to the new value
Converted[i,j] = value + (value * ((Change[0])/100))
#Convert back to 1D list
Converted.tolist()
return (Converted)
a = [16,2,20,4,14,6,70,8,9,100,32,15,7,14,50,20,17,10,9,20,7,17,50,2,19,20,21,22,23,24,25]
samples = [2, 7]
grid_details = [10,50,100]
result = grid(a,samples,grid_details)
print(result)
PS:我不想避免修改之前修改过的网格中的任何值，无论是主要值还是周围值。
最佳答案
首先，我不太确定你在问什么，如果我完全误解了你的问题，请原谅我......
你说你只想修改第一个等于给定值的项目，而不是所有的项目。如果是这样，您将需要添加 break在找到第一个值之后，否则您将继续循环并修改所有其他值。
但是，有更好的方法来做您想要的。
此外，您在顶部导入 numpy，然后从不(？)使用它...
这正是您想要使用 numpy 的那种事情，所以我将给出一个使用它的示例。
您似乎只是将函数应用于二维数组的 3x3 移动窗口，其中数组的值与某个给定值匹配。
如果我们想将给定索引周围的 3x3 区域设置为某个值，我们只需执行如下操作:
x[i-1:i+1, j-1:j+1] = value
...哪里x是你的数组，i和 j是行和列，value是您要将它们设置为的值。 (类似地，x[i-1:i+1, j-1:j+1] 返回围绕 <i,j> 的 3x3 数组)
此外，如果我们想知道 <i,j>指示特定值在数组中出现的位置，我们可以使用 numpy.argwhere 这将返回 <i,j> 的列表指示给定条件为真的每个地方。
(在 numpy 数组上使用条件会生成一个 bool 数组，显示条件为真或假。因此，x >= 10 将生成与 x 形状相同的 bool 数组，不是 简单地 True 或 False 。这让你可以做一些好的事情，比如 x[x>100] = 10 将 x 中所有大于 100 的值设置为 10。)
总而言之，我相信这段代码可以满足您的需求:
import numpy as np
# First let's generate some data and set a few duplicate values
data = np.arange(100).reshape(10,10)
data[9,9] = 2
data[8,6] = 53
print 'Original Data:'
print data
# We want to replace the _first_ occurences of "samples" with the corresponding
# value in "grid_details" within a 3x3 window...
samples = [2, 53, 69]
grid_details = [200,500,100]
nrows, ncols = data.shape
for value, new_value in zip(samples, grid_details):
# Notice that were're indexing the _first_ item than argwhere returns!
i,j = np.argwhere(data == value)[0]
# We need to make sure that we don't index outside the boundaries of the
# array (which would cause a "wraparound" assignment)
istart, istop = max(i-1, 0), min(i+2, nrows)
jstart, jstop = max(j-1, 0), min(j+2, ncols)
# Set the value within a 3x3 window to be "new_value"
data[istart:istop, jstart:jstop] = new_value
print 'Modified Data:'
print data
这会产生:
Original Data:
[[ 0 1 2 3 4 5 6 7 8 9]
[10 11 12 13 14 15 16 17 18 19]
[20 21 22 23 24 25 26 27 28 29]
[30 31 32 33 34 35 36 37 38 39]
[40 41 42 43 44 45 46 47 48 49]
[50 51 52 53 54 55 56 57 58 59]
[60 61 62 63 64 65 66 67 68 69]
[70 71 72 73 74 75 76 77 78 79]
[80 81 82 83 84 85 50 87 88 89]
[90 91 92 93 94 95 96 97 98 2]]
Modified Data:
[[ 0 200 200 200 4 5 6 7 8 9]
[ 10 200 200 200 14 15 16 17 18 19]
[ 20 21 22 23 24 25 26 27 28 29]
[ 30 31 32 33 34 35 36 37 38 39]
[ 40 41 500 500 500 45 46 47 48 49]
[ 50 51 500 500 500 55 56 57 100 100]
[ 60 61 500 500 500 65 66 67 100 100]
[ 70 71 72 73 74 75 76 77 100 100]
[ 80 81 82 83 84 85 50 87 88 89]
[ 90 91 92 93 94 95 96 97 98 2]]
最后，您提到您希望“将某物同时视为 N 维数组和“平面”列表”。从某种意义上说，这就是 numpy 数组的本来面目。
例如:
import numpy as np
x = np.arange(9)
y = x.reshape(3,3)
print x
print y
y[2,2] = 10000
print x
print y
在这里，y是 x 的“ View ” .如果我们更改 y 的元素我们更改x的对应元素反之亦然。
类似地，如果我们有一个 2D 数组(或 3D、4D 等)，我们希望将其视为“平面”1D 数组，您只需调用 flat_array = y.ravel()其中 y是你的二维数组。
无论如何，希望对您有所帮助!
关于Python 3.1-网格模拟概念问题，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/4870771/