blocks|key|2002845|text|什么是X？如果它是一个2D数组，那么如何将其行与一个数字进行比较：i+<+3|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2002846|在OP的评论后编辑：|2002847|A+=+array([[0,+1,+2],+[0,+2,+0]])
X+=+array([[0,+1,+2],+[1,+2,+0],+[2,+1,+2],+[3,+2,+0]])|code-block|syntax|javascript|2002848|将X中第一个元素为<+3的所有行添加到A|2002849|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]])|2002850|entityMap^0|3|1|X|5|0|0|0|1|1|9|3|J|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$]]

What is <code>X</code>? If it is a 2D-array, how can you then compare its row to a number: <code>i &lt; 3</code>?

EDIT after OP's comment:

<pre><code>A = array([[0, 1, 2], [0, 2, 0]])
X = array([[0, 1, 2], [1, 2, 0], [2, 1, 2], [3, 2, 0]])
</code></pre>

add to <code>A</code> all rows from <code>X</code> where the first element <code>&lt; 3</code>:

<pre><code>import numpy as np
A = np.vstack((A, X[X[:,0] &lt; 3]))

# returns: 
array([[0, 1, 2],
 [0, 2, 0],
 [0, 1, 2],
 [1, 2, 0],
 [2, 1, 2]])
</code></pre>

blocks|key|1789156|text|您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789157|newrow+=+[1,+2,+3]
A+=+numpy.vstack([A,+newrow])|code-block|syntax|javascript|1789158|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can do this:
<pre><code>newrow = [1, 2, 3]
A = numpy.vstack([A, newrow])
</code></pre>

blocks|key|2002868|text|您还可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2002869|newrow+=+[1,2,3]
A+=+numpy.concatenate((A,newrow))|code-block|syntax|javascript|2002870|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You can also do this:

<pre><code>newrow = [1,2,3]
A = numpy.concatenate((A,newrow))
</code></pre>

blocks|key|1789243|text|如果您可以在单个操作中完成构造，那么类似于vstack-with-+fine+indexing答案是一种很好的方法。但是，如果您的条件更复杂，或者您的行在运行时出现，您可能想要增加数组。事实上，做这样的事情的数字方式--动态增加一个数组--就是动态增加一个列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1789244|A+=+np.array([[1,2,3],[4,5,6]])
Alist+=+[r+for+r+in+A]
for+i+in+range(100):
++++newrow+=+np.arange(3)%2Bi
++++if+i%255:
++++++++Alist.append(newrow)
A+=+np.array(Alist)
del+Alist|code-block|syntax|javascript|1789245|列表针对这种访问模式进行了高度优化；在列表形式中，您没有方便的numpy多维索引，但只要您在追加，就很难比行数组的列表做得更好。|1789246|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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:

<pre><code>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
</code></pre>

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.

blocks|key|1789352|text|您可以使用numpy.append()将一行追加到numpty数组中，并在以后对矩阵进行整形。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1789353|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)|code-block|syntax|javascript|1789354|entityMap^0|5|E|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can use <code>numpy.append()</code> to append a row to numpty array and reshape to a matrix later on.

<pre><code>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)
</code></pre>

blocks|key|2003028|text|我使用更快的'np.vstack‘，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003029|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])|code-block|syntax|javascript|2003030|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I use 'np.vstack' which is faster, EX: 

<pre><code>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])
</code></pre>

blocks|key|2003162|text|如果每行之后都不需要计算，那么在python中添加行，然后转换为numpy会快得多。下面是使用python+3.6和numpy+1.14进行的定时测试，每次添加100行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2003163|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|code-block|syntax|javascript|2003164|因此，对于七年前的原始问题，简单的解决方案是在将行转换为numpy数组后使用vstack()添加新行。但更现实的解决方案应该考虑到vstack在这些情况下的糟糕性能。如果不需要在每次加法后对数组运行数据分析，最好将新行缓冲到python行列表(实际上是列表列表)，并在进行任何数据分析之前使用vstack()将它们作为一个组添加到numpy数组中。|2003165|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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:

<pre><code>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
</code></pre>

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.

blocks|key|2003169|text|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)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2003170|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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)
</code></pre>

blocks|key|1789610|text|我使用numpy.insert(arr,+i,+the_object_to_be_added,+axis)在第i行(axis=0)或列(axis=1)插入object_to_be_added|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1789611|import+numpy+as+np

a+=+np.array([[1,+2,+3],+[5,+4,+6]])
#+array([[1,+2,+3],
#++++++++[5,+4,+6]])

np.insert(a,+1,+[55,+66],+axis=1)
#+array([[+1,+55,++2,++3],
#++++++++[+5,+66,++4,++6]])

np.insert(a,+2,+[50,+60,+70],+axis=0)
#+array([[+1,++2,++3],
#++++++++[+5,++4,++6],
#++++++++[50,+60,+70]])|code-block|syntax|javascript|1789612|讨论太陈旧了，但我希望它能帮助一些人。|1789613|entityMap^0|3|1E|1L|8|1V|8|25|I|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

I use <code>numpy.insert(arr, i, the_object_to_be_added, axis)</code> in order to insert <code>object_to_be_added</code> at the i'th row<code>(axis=0)</code> or column<code>(axis=1)</code>
<pre><code>import numpy as np

a = np.array([[1, 2, 3], [5, 4, 6]])
# array([[1, 2, 3],
# [5, 4, 6]])

np.insert(a, 1, [55, 66], axis=1)
# array([[ 1, 55, 2, 3],
# [ 5, 66, 4, 6]])

np.insert(a, 2, [50, 60, 70], axis=0)
# array([[ 1, 2, 3],
# [ 5, 4, 6],
# [50, 60, 70]])
</code></pre>
Too old discussion, but I hope it helps someone.

How does one add rows to a numpy array?

I have an array A:

<pre><code>A = array([[0, 1, 2], [0, 2, 0]])
</code></pre>

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:

<pre><code>for i in X:
 if i[0] &lt; 3:
 A.append(i)
</code></pre>

Is there a numpythonic way to do the equivalent?

Thanks,
S ;-)

Numpy - add row to array

数据分析

如何向numpy数组添加行？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，我希望将行从另一个数组X添加到此数组中。Numpy数组没有像lists那样的“append”方法，至少看起来是这样。如果A和X是列表，我只会这样做：for i in X: ...

问Numpy -向数组添加行
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