[英]How do you extract a column from a multi-dimensional array?
有人知道如何从 Python 中的多维数组中提取一列吗?
>>> import numpy as np
>>> A = np.array([[1,2,3,4],[5,6,7,8]])
>>> A
array([[1, 2, 3, 4],
[5, 6, 7, 8]])
>>> A[:,2] # returns the third columm
array([3, 7])
See also: "numpy.arange" and "reshape" to allocate memory另见:“numpy.arange”和“reshape”分配内存
Example: (Allocating a array with shaping of matrix (3x4))示例:(分配矩阵形状为 (3x4) 的数组)
nrows = 3
ncols = 4
my_array = numpy.arange(nrows*ncols, dtype='double')
my_array = my_array.reshape(nrows, ncols)
Could it be that you're using a NumPy array ?可能是您使用的是NumPy 数组吗? Python has the array module, but that does not support multi-dimensional arrays. Python 有数组模块,但不支持多维数组。 Normal Python lists are single-dimensional too.普通的 Python 列表也是一维的。
However, if you have a simple two-dimensional list like this:但是,如果您有一个像这样的简单二维列表:
A = [[1,2,3,4],
[5,6,7,8]]
then you can extract a column like this:然后你可以像这样提取一列:
def column(matrix, i):
return [row[i] for row in matrix]
Extracting the second column (index 1):提取第二列(索引 1):
>>> column(A, 1)
[2, 6]
Or alternatively, simply:或者,简单地说:
>>> [row[1] for row in A]
[2, 6]
If you have an array like如果你有一个像
a = [[1, 2], [2, 3], [3, 4]]
Then you extract the first column like that:然后你像这样提取第一列:
[row[0] for row in a]
So the result looks like this:所以结果看起来像这样:
[1, 2, 3]
check it out!一探究竟!
a = [[1, 2], [2, 3], [3, 4]]
a2 = zip(*a)
a2[0]
it is the same thing as above except somehow it is neater the zip does the work but requires single arrays as arguments, the *a syntax unpacks the multidimensional array into single array arguments它与上面的内容相同,只是 zip 可以更简洁地完成工作,但需要单个数组作为参数,*a 语法将多维数组解包为单个数组参数
>>> x = arange(20).reshape(4,5)
>>> x array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
if you want the second column you can use如果你想要第二列,你可以使用
>>> x[:, 1]
array([ 1, 6, 11, 16])
If you have a two-dimensional array in Python (not numpy), you can extract all the columns like so,如果你在 Python 中有一个二维数组(不是 numpy),你可以像这样提取所有列,
data = [
['a', 1, 2],
['b', 3, 4],
['c', 5, 6]
]
columns = list(zip(*data))
print("column[0] = {}".format(columns[0]))
print("column[1] = {}".format(columns[1]))
print("column[2] = {}".format(columns[2]))
Executing this code will yield,执行此代码将产生,
>>> print("column[0] = {}".format(columns[0]))
column[0] = ('a', 'b', 'c')
>>> print("column[1] = {}".format(columns[1]))
column[1] = (1, 3, 5)
>>> print("column[2] = {}".format(columns[2]))
column[2] = (2, 4, 6)
def get_col(arr, col):
return map(lambda x : x[col], arr)
a = [[1,2,3,4], [5,6,7,8], [9,10,11,12],[13,14,15,16]]
print get_col(a, 3)
map function in Python is another way to go. Python 中的 map 函数是另一种方法。
array = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
col1 = [val[1] for val in array]
col2 = [val[2] for val in array]
col3 = [val[3] for val in array]
col4 = [val[4] for val in array]
print(col1)
print(col2)
print(col3)
print(col4)
Output:
[1, 5, 9, 13]
[2, 6, 10, 14]
[3, 7, 11, 15]
[4, 8, 12, 16]
[matrix[i][column] for i in range(len(matrix))]
The itemgetter operator can help too, if you like map-reduce style python, rather than list comprehensions, for a little variety! itemgetter 运算符也可以提供帮助,如果您喜欢 map-reduce 样式的 python,而不是列表推导式,则可以提供一些帮助!
# tested in 2.4
from operator import itemgetter
def column(matrix,i):
f = itemgetter(i)
return map(f,matrix)
M = [range(x,x+5) for x in range(10)]
assert column(M,1) == range(1,11)
You can use this as well:您也可以使用它:
values = np.array([[1,2,3],[4,5,6]])
values[...,0] # first column
#[1,4]
Note: This is not working for built-in array and not aligned (eg np.array([[1,2,3],[4,5,6,7]]) )注意:这不适用于内置数组且未对齐(例如 np.array([[1,2,3],[4,5,6,7]]) )
I think you want to extract a column from an array such as an array below我认为您想从数组中提取一列,例如下面的数组
import numpy as np
A = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])
Now if you want to get the third column in the format现在,如果您想获取格式的第三列
D=array[[3],
[7],
[11]]
Then you need to first make the array a matrix然后你需要首先使数组成为矩阵
B=np.asmatrix(A)
C=B[:,2]
D=asarray(C)
And now you can do element wise calculations much like you would do in excel.现在您可以像在 excel 中一样进行元素计算。
let's say we have n X m
matrix( n
rows and m
columns) say 5 rows and 4 columns假设我们有n X m
矩阵( n
行和m
列)说 5 行和 4 列
matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16],[17,18,19,20]]
To extract the columns in python, we can use list comprehension like this要在 python 中提取列,我们可以像这样使用列表推导
[ [row[i] for row in matrix] for in range(4) ]
You can replace 4 by whatever number of columns your matrix has.您可以将 4 替换为矩阵具有的任意数量的列。 The result is结果是
[ [1,5,9,13,17],[2,10,14,18],[3,7,11,15,19],[4,8,12,16,20] ]
One more way using matrices使用矩阵的另一种方法
>>> from numpy import matrix
>>> a = [ [1,2,3],[4,5,6],[7,8,9] ]
>>> matrix(a).transpose()[1].getA()[0]
array([2, 5, 8])
>>> matrix(a).transpose()[0].getA()[0]
array([1, 4, 7])
Just use transpose(), then you can get the columns as easy as you get rows只需使用 transpose(),就可以像获取行一样简单地获取列
matrix=np.array(originalMatrix).transpose()
print matrix[NumberOfColumns]
Well a 'bit' late ...好吧,有点晚了……
In case performance matters and your data is shaped rectangular, you might also store it in one dimension and access the columns by regular slicing eg ...如果性能很重要并且您的数据呈矩形,您也可以将其存储在一维中并通过常规切片访问列,例如...
A = [[1,2,3,4],[5,6,7,8]] #< assume this 4x2-matrix
B = reduce( operator.add, A ) #< get it one-dimensional
def column1d( matrix, dimX, colIdx ):
return matrix[colIdx::dimX]
def row1d( matrix, dimX, rowIdx ):
return matrix[rowIdx:rowIdx+dimX]
>>> column1d( B, 4, 1 )
[2, 6]
>>> row1d( B, 4, 1 )
[2, 3, 4, 5]
The neat thing is this is really fast.整洁的事情是这真的很快。 However , negative indexes don't work here!但是,负索引在这里不起作用! So you can't access the last column or row by index -1.所以你不能通过索引-1访问最后一列或最后一行。
If you need negative indexing you can tune the accessor-functions a bit, eg如果您需要负索引,您可以稍微调整访问器功能,例如
def column1d( matrix, dimX, colIdx ):
return matrix[colIdx % dimX::dimX]
def row1d( matrix, dimX, dimY, rowIdx ):
rowIdx = (rowIdx % dimY) * dimX
return matrix[rowIdx:rowIdx+dimX]
If you want to grab more than just one column just use slice:如果您想获取的不仅仅是一列,只需使用 slice:
a = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
print(a[:, [1, 2]])
[[2 3]
[5 6]
[8 9]]
Despite using zip(*iterable)
to transpose a nested list, you can also use the following if the nested lists vary in length:尽管使用zip(*iterable)
转置嵌套列表,但如果嵌套列表的长度不同,您也可以使用以下内容:
map(None, *[(1,2,3,), (4,5,), (6,)])
results in:结果是:
[(1, 4, 6), (2, 5, None), (3, None, None)]
The first column is thus:因此,第一列是:
map(None, *[(1,2,3,), (4,5,), (6,)])[0]
#>(1, 4, 6)
I prefer the next hint: having the matrix named matrix_a
and use column_number
, for example:我更喜欢下一个提示:将矩阵命名为matrix_a
并使用column_number
,例如:
import numpy as np
matrix_a = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])
column_number=2
# you can get the row from transposed matrix - it will be a column:
col=matrix_a.transpose()[column_number]
All columns from a matrix into a new list:矩阵中的所有列到一个新列表中:
N = len(matrix)
column_list = [ [matrix[row][column] for row in range(N)] for column in range(N) ]
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