entityMap|blocks|key|13ld7|text|给你。你只需要使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|8cp1u|(或者另选|dg0u3|)与广播相结合：|97c14|In+[6]:+data+-+vector[:,None]
Out[6]:
array([[0,+0,+0],
+++++++[0,+0,+0],
+++++++[0,+0,+0]])

In+[7]:+data+/+vector[:,None]
Out[7]:
array([[1,+1,+1],
+++++++[1,+1,+1],
+++++++[1,+1,+1]])|code-block|syntax|javascript^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|M|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|N|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|O|9|@]|A|@]|B|$J|K]]]]

Here you go. You just need to use <code>None</code> (or alternatively <code>np.newaxis</code>) combined with broadcasting:

<pre><code>In [6]: data - vector[:,None]
Out[6]:
array([[0, 0, 0],
 [0, 0, 0],
 [0, 0, 0]])

In [7]: data / vector[:,None]
Out[7]:
array([[1, 1, 1],
 [1, 1, 1],
 [1, 1, 1]])
</code></pre>

entityMap|blocks|key|d31tc|text|如前所述，使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9vsq8|或使用|277ef|是一个很好的方法来做到这一点。另一种选择是使用转置和广播，如|ddsen|(data.T+-+vector).T|code-block|syntax|javascript|cq2k3|和|bbrlj|(data.T+/+vector).T|8356s|对于更高维数的数组，您可能希望使用|258l5|NumPy数组或NumPy的方法|bnqid|函数。有很多方法可以做到这一点。|9g09g|有关广播的详细说明，请参阅|dgof3|http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html^0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Z|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|10|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|11|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|12|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|13|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|14|9|@]|A|@]|B|$J|K]]|$2|P|4|Q|6|7|8|15|9|@]|A|@]|B|$]]|$2|R|4|S|6|7|8|16|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|17|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|18|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|19|9|@]|A|@]|B|$]]]]

As has been mentioned, slicing with <code>None</code> or with <code>np.newaxes</code> is a great way to do this.
Another alternative is to use transposes and broadcasting, as in

<pre><code>(data.T - vector).T
</code></pre>

and

<pre><code>(data.T / vector).T
</code></pre>

For higher dimensional arrays you may want to use the <code>swapaxes</code> method of NumPy arrays or the NumPy <code>rollaxis</code> function.
There really are a lot of ways to do this.

For a fuller explanation of broadcasting, see
<a href="http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html">http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html</a>

entityMap|blocks|key|8g47k|text|JoshAdel的解决方案使用np.newaxis添加维度。另一种方法是使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|a0lbj|reshape()以对齐维度，为广播做准备|82k4b|..。|84qks|data+=+np.array([[1,1,1],[2,2,2],[3,3,3]])
vector+=+np.array([1,2,3])

data
#+array([[1,+1,+1],
#++++++++[2,+2,+2],
#++++++++[3,+3,+3]])
vector
#+array([1,+2,+3])

data.shape
#+(3,+3)
vector.shape
#+(3,)

data+/+vector.reshape((3,1))
#+array([[1,+1,+1],
#++++++++[1,+1,+1],
#++++++++[1,+1,+1]])|code-block|syntax|javascript|4kpnp|执行reshape()允许维度排列以进行广播：|3d2td|data:++++++++++++3+x+3
vector:++++++++++++++3
vector+reshaped:+3+x+1|a92j5|请注意|1qhkb|是可以的，但是它不会得到你想要的答案。它将每个|fo08p|列|75uvd|的|72prv|(而不是每个|e6hhf|行|e5ltk|)的每个相应元素|a8nbl|..。这就是如果你显式的重塑你会得到的结果|6jd7b|要成为|3ss7q|而不是|c3kfb|32oa4|data+/+vector
#+array([[1,+0,+0],
#++++++++[2,+1,+0],
#++++++++[3,+1,+1]])
data+/+vector.reshape((1,3))
#+array([[1,+0,+0],
#++++++++[2,+1,+0],
#++++++++[3,+1,+1]])^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|1C|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|1D|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|1E|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|1F|9|@]|A|@]|B|$J|K]]|$2|L|4|M|6|7|8|1G|9|@]|A|@]|B|$]]|$2|N|4|O|6|I|8|1H|9|@]|A|@]|B|$J|K]]|$2|P|4|Q|6|7|8|1I|9|@]|A|@]|B|$]]|$2|R|4|S|6|7|8|1J|9|@]|A|@]|B|$]]|$2|T|4|U|6|7|8|1K|9|@]|A|@]|B|$]]|$2|V|4|W|6|7|8|1L|9|@]|A|@]|B|$]]|$2|X|4|Y|6|7|8|1M|9|@]|A|@]|B|$]]|$2|Z|4|10|6|7|8|1N|9|@]|A|@]|B|$]]|$2|11|4|12|6|7|8|1O|9|@]|A|@]|B|$]]|$2|13|4|14|6|7|8|1P|9|@]|A|@]|B|$]]|$2|15|4|16|6|7|8|1Q|9|@]|A|@]|B|$]]|$2|17|4|18|6|7|8|1R|9|@]|A|@]|B|$]]|$2|19|4|F|6|7|8|1S|9|@]|A|@]|B|$]]|$2|1A|4|1B|6|I|8|1T|9|@]|A|@]|B|$J|K]]]]

JoshAdel's solution uses np.newaxis to add a dimension. An alternative is to use <a href="http://docs.scipy.org/doc/numpy/user/basics.broadcasting.html" rel="nofollow">reshape() to align the dimensions in preparation for broadcasting</a>.

<pre><code>data = np.array([[1,1,1],[2,2,2],[3,3,3]])
vector = np.array([1,2,3])

data
# array([[1, 1, 1],
# [2, 2, 2],
# [3, 3, 3]])
vector
# array([1, 2, 3])

data.shape
# (3, 3)
vector.shape
# (3,)

data / vector.reshape((3,1))
# array([[1, 1, 1],
# [1, 1, 1],
# [1, 1, 1]])
</code></pre>

Performing the reshape() allows the dimensions to line up for broadcasting:

<pre><code>data: 3 x 3
vector: 3
vector reshaped: 3 x 1
</code></pre>

Note that <code>data/vector</code> is ok, but it doesn't get you the answer that you want. It divides each column of <code>array</code> (instead of each row) by each corresponding element of <code>vector</code>. It's what you would get if you explicitly reshaped <code>vector</code> to be <code>1x3</code> instead of <code>3x1</code>.

<pre><code>data / vector
# array([[1, 0, 0],
# [2, 1, 0],
# [3, 1, 1]])
data / vector.reshape((1,3))
# array([[1, 0, 0],
# [2, 1, 0],
# [3, 1, 1]])
</code></pre>

entityMap|blocks|key|f51k8|text|Pythonic式的方法是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|d6b4|np.divide(data.T,vector).T|code-block|syntax|javascript|20frp|这将负责整形，并且结果也是浮点格式的。在其他答案中，结果是四舍五入的整数格式。|apsed|#注意：|18vmr|数据和向量中的列数应匹配^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|N|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|O|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|P|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|Q|9|@]|A|@]|B|$]]|$2|L|4|M|6|7|8|R|9|@]|A|@]|B|$]]]]

Pythonic way to do this is ...

<pre><code>np.divide(data.T,vector).T
</code></pre>

This takes care of reshaping and also the results are in floating point format.
In other answers results are in rounded integer format.

#NOTE: No of columns in both data and vector should match

entityMap|blocks|key|7pkti|text|除了stackoverflowuser2010的答案之外，在一般情况下，您可以只使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2eb6n|data+=+np.array([[1,1,1],[2,2,2],[3,3,3]])

vector+=+np.array([1,2,3])

data+/+vector.reshape(-1,1)|code-block|syntax|javascript|19vs5|这将把你的向量变成一个|eukik|..。允许你随心所欲地进行元素级的操作。至少对我来说，这是最直观的方式，因为(在大多数情况下)+numpy只会使用相同内存的视图进行重塑，所以效率也很高。^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|7|8|O|9|@]|A|@]|B|$]]]]

Adding to the answer of stackoverflowuser2010, in the general case you can just use

<pre><code>data = np.array([[1,1,1],[2,2,2],[3,3,3]])

vector = np.array([1,2,3])

data / vector.reshape(-1,1)
</code></pre>

This will turn your vector into a <code>column matrix/vector</code>. Allowing you to do the elementwise operations as you wish. At least to me, this is the most intuitive way going about it and since (in most cases) numpy will just use a view of the same internal memory for the reshaping it's efficient too.

entityMap|blocks|key|67q0t|text|关键是重塑大小为(3，)到(3，1)的向量:将每行除以一个元素，或(1,3)：将每列除以一个元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|ck5qe|In[1]:+data/vector.reshape(-1,1)
Out[1]:
array([[1.,+1.,+1.],
+++++++[1.,+1.,+1.],
+++++++[1.,+1.,+1.]])

In[2]:+data/vector.reshape(1,-1)
Out[2]:
array([[1.++++++++,+0.5+++++++,+0.33333333],
+++++++[2.++++++++,+1.++++++++,+0.66666667],
+++++++[3.++++++++,+1.5+++++++,+1.++++++++]])|code-block|syntax|javascript|ag5sa|类似的：|8mu87|x+=+np.arange(9).reshape(3,3)
x
array([[0,+1,+2],
+++++++[3,+4,+5],
+++++++[6,+7,+8]])

x/np.sum(x,+axis=0,+keepdims=True)
array([[0.++++++++,+0.08333333,+0.13333333],
+++++++[0.33333333,+0.33333333,+0.33333333],
+++++++[0.66666667,+0.58333333,+0.53333333]])

x/np.sum(x,+axis=1,+keepdims=True)
array([[0.++++++++,+0.33333333,+0.66666667],
+++++++[0.25++++++,+0.33333333,+0.41666667],
+++++++[0.28571429,+0.33333333,+0.38095238]])

print(np.sum(x,+axis=0).shape)
print(np.sum(x,+axis=1).shape)
print(np.sum(x,+axis=0,+keepdims=True).shape)
print(np.sum(x,+axis=1,+keepdims=True).shape)
(3,)
(3,)
(1,+3)
(3,+1)^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|M|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|N|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|O|9|@]|A|@]|B|$F|G]]]]

The key is to reshape the vector of size (3,) to (3,1): divide each row by an element or (1,3): divide each column by an element. As data.shape does not correspond to vector.shape, NumPy automatically expands vector's shape to (3,3) and performs division, element-wise.
<pre class="lang-py prettyprint-override"><code>In[1]: data/vector.reshape(-1,1)
Out[1]:
array([[1., 1., 1.],
 [1., 1., 1.],
 [1., 1., 1.]])

In[2]: data/vector.reshape(1,-1)
Out[2]:
array([[1. , 0.5 , 0.33333333],
 [2. , 1. , 0.66666667],
 [3. , 1.5 , 1. ]])

</code></pre>
Similar:
<pre class="lang-py prettyprint-override"><code>x = np.arange(9).reshape(3,3)
x
array([[0, 1, 2],
 [3, 4, 5],
 [6, 7, 8]])

x/np.sum(x, axis=0, keepdims=True)
array([[0. , 0.08333333, 0.13333333],
 [0.33333333, 0.33333333, 0.33333333],
 [0.66666667, 0.58333333, 0.53333333]])

x/np.sum(x, axis=1, keepdims=True)
array([[0. , 0.33333333, 0.66666667],
 [0.25 , 0.33333333, 0.41666667],
 [0.28571429, 0.33333333, 0.38095238]])

print(np.sum(x, axis=0).shape)
print(np.sum(x, axis=1).shape)
print(np.sum(x, axis=0, keepdims=True).shape)
print(np.sum(x, axis=1, keepdims=True).shape)
(3,)
(3,)
(1, 3)
(3, 1)
</code></pre>

Suppose I have a numpy array:

<pre><code>data = np.array([[1,1,1],[2,2,2],[3,3,3]])
</code></pre>

and I have a corresponding "vector:"

<pre><code>vector = np.array([1,2,3])
</code></pre>

How do I operate on <code>data</code> along each row to either subtract or divide so the result is:

<pre><code>sub_result = [[0,0,0], [0,0,0], [0,0,0]]
div_result = [[1,1,1], [1,1,1], [1,1,1]]
</code></pre>

Long story short: How do I perform an operation on each row of a 2D array with a 1D array of scalars that correspond to each row?

Numpy: Divide each row by a vector element

 假设我有一个numpy数组： data = np.array([[1,1,1],[2,2,2],[3,3,3]]) 我有一个对应的“向量”： vector = np.array([1,2,3]) 如何操作 沿着每一行进行减法或除法，因此结果为： sub_result = [[0,0,0], [0,0,0], [0,0...

问Numpy:将每行除以一个向量元素
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