如何在python中向3维数组添加元素
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How to add elements to 3 dimensional array in python
提问by lokesh
I am trying to store data in three-dimensional array i.e, x[0][0][0]in Python. How to initialize x, and add values to it? I have tried this:
我正在尝试将数据存储在三维数组中，即x[0][0][0]在 Python 中。如何初始化x，并为其添加值？我试过这个：
x=[]
x[0][0][0]=value1
x[0][0].append(value1)
both lines are giving out of range error. How to do it? I want it like: x[0][0][0]=value1, x[1][0][0]=value2, x[0][1][0]=value3etc. How to achieve this in Python?
两条线都给出了超出范围的错误。怎么做？我想它像：x[0][0][0]=value1，x[1][0][0]=value2，x[0][1][0]=value3等。如何在Python实现这一目标？
I am looking to generate this kind of array:
我正在寻找生成这种数组：
x=[[[11,[111],[112]],[12],[13]],[[21,[211],[212]],[22],[23],[24]],[[31],[32]]]
x[0][0][0] will give 11
x[1][0][0] 21
x[0][0][1] 111
etc.
等等。
采纳答案by amaurea
I recommend using numpyfor multidimensional arrays. It makes it much more convenient, and much faster. This would look like:
我推荐numpy用于多维数组。它使它更方便，更快。这看起来像：
import numpy as np
x = np.zeros((10,20,30)) # Make a 10 by 20 by 30 array
x[0,0,0] = value1
Still, if you don't want to use numpy, or need non-rectangular multi-dimensional arrays, you will need to treat it as a list of lists of lists, and initialize each list:
尽管如此，如果您不想使用numpy，或需要非矩形多维数组，则需要将其视为列表列表的列表，并初始化每个列表：
x = []
x.append([])
x[0].append([])
x[0][0].append(value1)
Edit: Or you could use the compact notation shown in ndpu's answer (x = [[[value1]]]).
编辑：或者您可以使用 ndpu 的答案 ( x = [[[value1]]]) 中显示的紧凑符号。
回答by Andrew Walker
If you can use numpy, you can initialize a fixed size array as:
如果可以使用 numpy，则可以将固定大小的数组初始化为：
import numpy
x = numpy.zeros((i, j, k))
where i, j and k are the dimensions required.
其中 i、j 和 k 是所需的维度。
You can then index into that array using slice notation:
然后，您可以使用切片表示法对该数组进行索引：
x[0, 0, 0] = value1
x[1, 0, 0] = value2
回答by ndpu
>>> x=[[[[]]]]
>>> x[0][0][0]=0
>>> x
[[[0]]]
>>> x[0][0].append(1)
>>> x
[[[0, 1]]]
回答by HYRY
If you are creating some 3D sparse array, you can save all the data in a dict:
如果您正在创建一些 3D 稀疏数组，您可以将所有数据保存在一个 dict 中：
x={}
x[0,0,0] = 11
x[1,0,0] = 21
x[0,1,1] = 111
or:
或者：
from collections import defaultdict
x = defaultdict(lambda :defaultdict(lambda :defaultdict(int)))
x[0][0][0] = 11
x[1][0][0] = 21
x[0][0][1] = 111
回答by user2099484
A compilation of techniques above with respect to effectively creating a two-dimensional dictis:
上面关于有效创建二维字典的技术汇编是：
from collections import defaultdict
x = defaultdict(lambda :defaultdict())
x["a"]["b"] = 123
x["a"]["c"] = 234
x["b"]["a"] = 234
x["b"]["c"] = 234
x["c"]["a"] = 234
x["c"]["b"] = 234
for i in x:
for j in x[i]: print i, j, x[i][j]
Produces:
产生：
a c 234
a b 123
c a 234
c b 234
b a 234
b c 234
To increase the number of dimensions (e.g., to three-dimensions), simply increase the lambda "nest" as HYRY shows:
要增加维数（例如，增加到三个维度），只需增加 lambda“嵌套”，如 HYRY 所示：
x = defaultdict(lambda :defaultdict(lambda :defaultdict(int)))
回答by Peter Hausdorff Smith K.
Interator. I use function lambda for create matrix with [ [ 0 for j in range(n)] for i in range(m) ] in python 3. In the version 2 a used map functions. In this moment think like use array module (pure python) to create matrixes.
交互器。我使用函数 lambda 在 python 3 中使用 [ [ 0 for j in range(n)] for i in range(m) ] 创建矩阵。在版本 2 中使用了映射函数。在这一刻想像使用数组模块（纯python）来创建矩阵。
>>> arr = lambda m,n,l : [ [ [0 for k in range(l)] for j in range(n)] for i in range(m) ]
>>> m = arr(2,3,4)
>>> m
[[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]], [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]]
回答by Wayne Workman
I just came up with this, it's more dynamic and simple.
我刚想出这个，它更动态和简单。
# Define how large to make the object.
size = 3
# Build the three dimensional list.
memory = []
for x in range(0,size):
memory.append([])
for y in range(0,size):
memory[x].append([])
for z in range(0,size):
memory[x][y].append(0) # Fill with zeros.
# Iterate through all values.
for x in range(0,size):
for y in range(0,size):
for z in range(0,size):
print 'memory[' + str(x) + '][' + str(y) + '][' + str(z) + ']=' + str(memory[x][y][z])
# Example access.
print 'Example access:'
print 'memory[0][1][2]=' + str(memory[0][1][2])
Output:
输出：
memory[0][0][0]=0
memory[0][0][1]=0
memory[0][0][2]=0
memory[0][1][0]=0
memory[0][1][1]=0
memory[0][1][2]=0
memory[0][2][0]=0
memory[0][2][1]=0
memory[0][2][2]=0
memory[1][0][0]=0
memory[1][0][1]=0
memory[1][0][2]=0
memory[1][1][0]=0
memory[1][1][1]=0
memory[1][1][2]=0
memory[1][2][0]=0
memory[1][2][1]=0
memory[1][2][2]=0
memory[2][0][0]=0
memory[2][0][1]=0
memory[2][0][2]=0
memory[2][1][0]=0
memory[2][1][1]=0
memory[2][1][2]=0
memory[2][2][0]=0
memory[2][2][1]=0
memory[2][2][2]=0
Example access:
memory[0][1][2]=0
