问题:numpy-将行添加到数组
如何将行添加到numpy数组?
我有一个数组A:
A = array([[0, 1, 2], [0, 2, 0]])
如果X中每行的第一个元素满足特定条件,我希望从另一个数组X向该数组添加行。
Numpy数组没有像列表那样的“追加”方法,或者看起来。
如果A和X是列表,我只会这样做:
for i in X:
if i[0] < 3:
A.append(i)
是否有numpythonic的方式来做等效的?
谢谢,S ;-)
How does one add rows to a numpy array?
I have an array A:
A = array([[0, 1, 2], [0, 2, 0]])
I wish to add rows to this array from another array X if the first element of each row in X meets a specific condition.
Numpy arrays do not have a method ‘append’ like that of lists, or so it seems.
If A and X were lists I would merely do:
for i in X:
if i[0] < 3:
A.append(i)
Is there a numpythonic way to do the equivalent?
Thanks,
S ;-)
回答 0
什么X
啊 如果它是一个二维数组,你怎么能那么其行比作一个号码:i < 3
?
OP评论后编辑:
A = array([[0, 1, 2], [0, 2, 0]])
X = array([[0, 1, 2], [1, 2, 0], [2, 1, 2], [3, 2, 0]])
A
从X
第一个元素添加到所有行< 3
:
import numpy as np
A = np.vstack((A, X[X[:,0] < 3]))
# returns:
array([[0, 1, 2],
[0, 2, 0],
[0, 1, 2],
[1, 2, 0],
[2, 1, 2]])
What is X
? If it is a 2D-array, how can you then compare its row to a number: i < 3
?
EDIT after OP’s comment:
A = array([[0, 1, 2], [0, 2, 0]])
X = array([[0, 1, 2], [1, 2, 0], [2, 1, 2], [3, 2, 0]])
add to A
all rows from X
where the first element < 3
:
import numpy as np
A = np.vstack((A, X[X[:,0] < 3]))
# returns:
array([[0, 1, 2],
[0, 2, 0],
[0, 1, 2],
[1, 2, 0],
[2, 1, 2]])
回答 1
好吧,你可以这样做:
newrow = [1,2,3]
A = numpy.vstack([A, newrow])
well u can do this :
newrow = [1,2,3]
A = numpy.vstack([A, newrow])
回答 2
由于这个问题已经存在了7年,所以我使用的最新版本是numpy版本1.13和python3,我在向矩阵中添加一行时也做同样的事情,请记住在第二个参数中加上双括号,否则会引起尺寸误差。
在这里我要添加矩阵A
1 2 3
4 5 6
连续
7 8 9
相同的用法 np.r_
A= [[1, 2, 3], [4, 5, 6]]
np.append(A, [[7, 8, 9]], axis=0)
>> array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
#or
np.r_[A,[[7,8,9]]]
只是对某人感兴趣,如果您想添加一列,
array = np.c_[A,np.zeros(#A's row size)]
按照我们之前在矩阵A上所做的操作,向其中添加一列
np.c_[A, [2,8]]
>> array([[1, 2, 3, 2],
[4, 5, 6, 8]])
As this question is been 7 years before, in the latest version which I am using is numpy version 1.13, and python3, I am doing the same thing with adding a row to a matrix, remember to put a double bracket to the second argument, otherwise, it will raise dimension error.
In here I am adding on matrix A
1 2 3
4 5 6
with a row
7 8 9
same usage in np.r_
A= [[1, 2, 3], [4, 5, 6]]
np.append(A, [[7, 8, 9]], axis=0)
>> array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
#or
np.r_[A,[[7,8,9]]]
Just to someone’s intersted, if you would like to add a column,
array = np.c_[A,np.zeros(#A's row size)]
following what we did before on matrix A, adding a column to it
np.c_[A, [2,8]]
>> array([[1, 2, 3, 2],
[4, 5, 6, 8]])
回答 3
您也可以这样做:
newrow = [1,2,3]
A = numpy.concatenate((A,newrow))
You can also do this:
newrow = [1,2,3]
A = numpy.concatenate((A,newrow))
回答 4
如果每行之后都不需要进行计算,则在python中添加行然后转换为numpy会更快。以下是使用python 3.6与numpy 1.14进行的时序测试,添加了100行,一次添加一行:
import numpy as np
from time import perf_counter, sleep
def time_it():
# Compare performance of two methods for adding rows to numpy array
py_array = [[0, 1, 2], [0, 2, 0]]
py_row = [4, 5, 6]
numpy_array = np.array(py_array)
numpy_row = np.array([4,5,6])
n_loops = 100
start_clock = perf_counter()
for count in range(0, n_loops):
numpy_array = np.vstack([numpy_array, numpy_row]) # 5.8 micros
duration = perf_counter() - start_clock
print('numpy 1.14 takes {:.3f} micros per row'.format(duration * 1e6 / n_loops))
start_clock = perf_counter()
for count in range(0, n_loops):
py_array.append(py_row) # .15 micros
numpy_array = np.array(py_array) # 43.9 micros
duration = perf_counter() - start_clock
print('python 3.6 takes {:.3f} micros per row'.format(duration * 1e6 / n_loops))
sleep(15)
#time_it() prints:
numpy 1.14 takes 5.971 micros per row
python 3.6 takes 0.694 micros per row
因此,七年前对原始问题的简单解决方案是在将行转换为numpy数组后,使用vstack()添加新行。但是更现实的解决方案应该考虑在这些情况下vstack的性能不佳。如果您不需要在每次添加后对数组进行数据分析,最好将新行缓冲到python行列表(实际上是列表列表)中,然后将它们作为一个组添加到numpy数组中在进行任何数据分析之前使用vstack()。
If no calculations are necessary after every row, it’s much quicker to add rows in python, then convert to numpy. Here are timing tests using python 3.6 vs. numpy 1.14, adding 100 rows, one at a time:
import numpy as np
from time import perf_counter, sleep
def time_it():
# Compare performance of two methods for adding rows to numpy array
py_array = [[0, 1, 2], [0, 2, 0]]
py_row = [4, 5, 6]
numpy_array = np.array(py_array)
numpy_row = np.array([4,5,6])
n_loops = 100
start_clock = perf_counter()
for count in range(0, n_loops):
numpy_array = np.vstack([numpy_array, numpy_row]) # 5.8 micros
duration = perf_counter() - start_clock
print('numpy 1.14 takes {:.3f} micros per row'.format(duration * 1e6 / n_loops))
start_clock = perf_counter()
for count in range(0, n_loops):
py_array.append(py_row) # .15 micros
numpy_array = np.array(py_array) # 43.9 micros
duration = perf_counter() - start_clock
print('python 3.6 takes {:.3f} micros per row'.format(duration * 1e6 / n_loops))
sleep(15)
#time_it() prints:
numpy 1.14 takes 5.971 micros per row
python 3.6 takes 0.694 micros per row
So, the simple solution to the original question, from seven years ago, is to use vstack() to add a new row after converting the row to a numpy array. But a more realistic solution should consider vstack’s poor performance under those circumstances. If you don’t need to run data analysis on the array after every addition, it is better to buffer the new rows to a python list of rows (a list of lists, really), and add them as a group to the numpy array using vstack() before doing any data analysis.
回答 5
import numpy as np
array_ = np.array([[1,2,3]])
add_row = np.array([[4,5,6]])
array_ = np.concatenate((array_, add_row), axis=0)
import numpy as np
array_ = np.array([[1,2,3]])
add_row = np.array([[4,5,6]])
array_ = np.concatenate((array_, add_row), axis=0)
回答 6
如果您可以在一个操作中完成构造,那么类似vstack-with-fancy-indexing的答案就是很好的方法。但是,如果您的情况更加复杂,或者您的行不断增加,那么您可能想要增加数组。实际上,执行类似这样的numpythonic方法-动态增长数组-是动态增长列表:
A = np.array([[1,2,3],[4,5,6]])
Alist = [r for r in A]
for i in range(100):
newrow = np.arange(3)+i
if i%5:
Alist.append(newrow)
A = np.array(Alist)
del Alist
列表针对这种访问模式进行了高度优化。在列表形式时,您没有方便的numpy多维索引,但是只要您要追加,就很难比行数组列表做得更好。
If you can do the construction in a single operation, then something like the vstack-with-fancy-indexing answer is a fine approach. But if your condition is more complicated or your rows come in on the fly, you may want to grow the array. In fact the numpythonic way to do something like this – dynamically grow an array – is to dynamically grow a list:
A = np.array([[1,2,3],[4,5,6]])
Alist = [r for r in A]
for i in range(100):
newrow = np.arange(3)+i
if i%5:
Alist.append(newrow)
A = np.array(Alist)
del Alist
Lists are highly optimized for this kind of access pattern; you don’t have convenient numpy multidimensional indexing while in list form, but for as long as you’re appending it’s hard to do better than a list of row arrays.
回答 7
我使用更快的“ np.vstack”,例如:
import numpy as np
input_array=np.array([1,2,3])
new_row= np.array([4,5,6])
new_array=np.vstack([input_array, new_row])
I use ‘np.vstack’ which is faster, EX:
import numpy as np
input_array=np.array([1,2,3])
new_row= np.array([4,5,6])
new_array=np.vstack([input_array, new_row])
回答 8
您可以用来numpy.append()
在numpty数组后附加一行,然后再将其整形为矩阵。
import numpy as np
a = np.array([1,2])
a = np.append(a, [3,4])
print a
# [1,2,3,4]
# in your example
A = [1,2]
for row in X:
A = np.append(A, row)
You can use numpy.append()
to append a row to numpty array and reshape to a matrix later on.
import numpy as np
a = np.array([1,2])
a = np.append(a, [3,4])
print a
# [1,2,3,4]
# in your example
A = [1,2]
for row in X:
A = np.append(A, row)