如何从多维数组中提取列?

问题:如何从多维数组中提取列?

有人知道如何在Python中从多维数组中提取列吗?

Does anybody know how to extract a column from a multi-dimensional array in Python?


回答 0

>>> 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])

另请参阅:“ numpy.arange”和“ reshape”以分配内存

示例:(使用矩阵整形(3×4)分配数组)

nrows = 3
ncols = 4
my_array = numpy.arange(nrows*ncols, dtype='double')
my_array = my_array.reshape(nrows, ncols)
>>> 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

Example: (Allocating a array with shaping of matrix (3×4))

nrows = 3
ncols = 4
my_array = numpy.arange(nrows*ncols, dtype='double')
my_array = my_array.reshape(nrows, ncols)

回答 1

可能是您正在使用NumPy数组吗?Python具有数组模块,但是不支持多维数组。普通的Python列表也是一维的。

但是,如果您有一个简单的二维列表,例如:

A = [[1,2,3,4],
     [5,6,7,8]]

然后您可以提取一个像这样的列:

def column(matrix, i):
    return [row[i] for row in matrix]

提取第二列(索引1):

>>> column(A, 1)
[2, 6]

或者,简单地:

>>> [row[1] for row in A]
[2, 6]

Could it be that you’re using a NumPy array? Python has the array module, but that does not support multi-dimensional arrays. Normal Python lists are single-dimensional too.

However, if you have a simple two-dimensional list like this:

A = [[1,2,3,4],
     [5,6,7,8]]

then you can extract a column like this:

def column(matrix, i):
    return [row[i] for row in matrix]

Extracting the second column (index 1):

>>> column(A, 1)
[2, 6]

Or alternatively, simply:

>>> [row[1] for row in A]
[2, 6]

回答 2

如果你有一个像

a = [[1, 2], [2, 3], [3, 4]]

然后像这样提取第一列:

[row[0] for row in a]

所以结果看起来像这样:

[1, 2, 3]

If you have an array like

a = [[1, 2], [2, 3], [3, 4]]

Then you extract the first column like that:

[row[0] for row in a]

So the result looks like this:

[1, 2, 3]

回答 3

看看这个!

a = [[1, 2], [2, 3], [3, 4]]
a2 = zip(*a)
a2[0]

它与上面相同,只是它使zip工作更整洁,但需要单个数组作为参数,* a语法将多维数组解压缩为单个数组参数

check it out!

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


回答 4

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)

Python中的map函数是另一种方法。

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.


回答 5

>>> 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]])

如果您想使用第二栏,可以使用

>>> x[:, 1]
array([ 1,  6, 11, 16])
>>> 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

>>> x[:, 1]
array([ 1,  6, 11, 16])

回答 6

[matrix[i][column] for i in range(len(matrix))]
[matrix[i][column] for i in range(len(matrix))]

回答 7

如果您喜欢map-reduce样式的python,而不是列表推导,itemgetter运算符也可以提供帮助!

# 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)

The itemgetter operator can help too, if you like map-reduce style python, rather than list comprehensions, for a little variety!

# 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)

回答 8

您也可以使用此:

values = np.array([[1,2,3],[4,5,6]])
values[...,0] # first column
#[1,4]

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]]))

You can use this as well:

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 (e.g. np.array([[1,2,3],[4,5,6,7]]) )


回答 9

我想您想从数组(例如下面的数组)中提取列

import numpy as np
A = np.array([[1,2,3,4],[5,6,7,8],[9,10,11,12]])

现在,如果要获取格式的第三列

D=array[[3],
[7],
[11]]

然后,您需要首先使数组成为矩阵

B=np.asmatrix(A)
C=B[:,2]
D=asarray(C)

现在,您可以像在excel中一样进行元素明智的计算。

I think you want to extract a column from an array such as an array below

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

D=array[[3],
[7],
[11]]

Then you need to first make the array a matrix

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.


回答 10

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]]

要提取python中的列,我们可以像这样使用列表理解

[ [row[i] for row in matrix] for in range(4) ]

您可以将矩阵中的列数替换为4。结果是

[ [1,5,9,13,17],[2,10,14,18],[3,7,11,15,19],[4,8,12,16,20] ]

let’s say we have n X m matrix(n rows and m columns) say 5 rows and 4 columns

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

[ [row[i] for row in matrix] for in range(4) ]

You can replace 4 by whatever number of columns your matrix has. The result is

[ [1,5,9,13,17],[2,10,14,18],[3,7,11,15,19],[4,8,12,16,20] ]


回答 11

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]
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]

回答 12

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])

One more way using matrices

>>> 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])

回答 13

如果您在Python中有一个二维数组(不是numpy),则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]))

执行此代码会产生

>>> 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)

当然,您可以按索引提取单个列(例如columns[0]

If you have a two-dimensional array in Python (not numpy), you can extract all the columns like so,

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,

>>> 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)

Of course, you can extract a single column by index (e.g. columns[0])


回答 14

尽管使用zip(*iterable)了转置嵌套列表,但是如果嵌套列表的长度不同,也可以使用以下内容:

map(None, *[(1,2,3,), (4,5,), (6,)])

结果是:

[(1, 4, 6), (2, 5, None), (3, None, None)]

因此,第一列是:

map(None, *[(1,2,3,), (4,5,), (6,)])[0]
#>(1, 4, 6)

Despite using zip(*iterable) to transpose a nested list, you can also use the following if the nested lists vary in length:

map(None, *[(1,2,3,), (4,5,), (6,)])

results in:

[(1, 4, 6), (2, 5, None), (3, None, None)]

The first column is thus:

map(None, *[(1,2,3,), (4,5,), (6,)])[0]
#>(1, 4, 6)

回答 15

好吧,有点晚…

如果性能很重要并且您的数据呈矩形,则也可以将其存储为一维,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]

整洁的是,这真的很快。但是,负索引在这里不起作用!因此,您无法按索引-1访问最后一列或最后一行。

如果需要负索引,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]

Well a ‘bit’ late …

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 e.g. …

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. However, negative indexes don’t work here! So you can’t access the last column or row by index -1.

If you need negative indexing you can tune the accessor-functions a bit, e.g.

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]

回答 16

如果要获取不止一列,请使用slice:

 a = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
    print(a[:, [1, 2]])
[[2 3]
[5 6]
[8 9]]

If you want to grab more than just one column just use slice:

 a = np.array([[1, 2, 3],[4, 5, 6],[7, 8, 9]])
    print(a[:, [1, 2]])
[[2 3]
[5 6]
[8 9]]

回答 17

我更喜欢下一条提示:将矩阵命名为matrix_ause 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]

I prefer the next hint: having the matrix named matrix_a and use column_number, for example:

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]

回答 18

只需使用transpose(),然后您就可以像获得行一样轻松获得列

matrix=np.array(originalMatrix).transpose()
print matrix[NumberOfColum]

Just use transpose(), then you can get the colummns as easy as you get rows

matrix=np.array(originalMatrix).transpose()
print matrix[NumberOfColum]

回答 19

矩阵中的所有列到新列表中:

N = len(matrix) 
column_list = [ [matrix[row][column] for row in range(N)] for column in range(N) ]

All columns from a matrix into a new list:

N = len(matrix) 
column_list = [ [matrix[row][column] for row in range(N)] for column in range(N) ]