blocks|key|4476284|text|这将是很好的，如果以某种矢量化的方式，但我仍然是一个新手，所以这是我现在能想到的一切！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4476285|import+numpy+as+np,numba+as+nb
a=np.array([[1,+2,+3,+4],
+++++++++++++++++[2,+3,+1],
+++++++++++++++++[5,+5,+5,+5,5],
+++++++++++++++++[1,+1]])
@nb.jit()
def+f(a):
++++l=len(max(a,key=len))
++++a0=np.empty(a.shape%2B(l,))
++++for+n,i+in+enumerate(a.flat):
++++++++a0[n]=np.pad(i,(0,l-len(i)),mode='constant')
++++a=a0
++++return+a

print(f(a))|code-block|syntax|javascript|4476286|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|$]]

This would be nice if in some vectorized way, but Im still a NOOB, so its all I could think now!

<pre><code>import numpy as np,numba as nb
a=np.array([[1, 2, 3, 4],
 [2, 3, 1],
 [5, 5, 5, 5,5],
 [1, 1]])
@nb.jit()
def f(a):
 l=len(max(a,key=len))
 a0=np.empty(a.shape+(l,))
 for n,i in enumerate(a.flat):
 a0[n]=np.pad(i,(0,l-len(i)),mode='constant')
 a=a0
 return a

print(f(a))
</code></pre>

blocks|key|4475538|text|这可能是一种方法-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4475539|def+numpy_fillna(data):
++++#+Get+lengths+of+each+row+of+data
++++lens+=+np.array([len(i)+for+i+in+data])

++++#+Mask+of+valid+places+in+each+row
++++mask+=+np.arange(lens.max())+<+lens[:,None]

++++#+Setup+output+array+and+put+elements+from+data+into+masked+positions
++++out+=+np.zeros(mask.shape,+dtype=data.dtype)
++++out[mask]+=+np.concatenate(data)
++++return+out|code-block|syntax|javascript|4475540|样本输入，输出-|4475541|In+[222]:+#+Input+object+dtype+array
+++++...:+data+=+np.array([[1,+2,+3,+4],
+++++...:++++++++++++++++++[2,+3,+1],
+++++...:++++++++++++++++++[5,+5,+5,+5,+8+,9+,5],
+++++...:++++++++++++++++++[1,+1]])

In+[223]:+numpy_fillna(data)
Out[223]:+
array([[1,+2,+3,+4,+0,+0,+0],
+++++++[2,+3,+1,+0,+0,+0,+0],
+++++++[5,+5,+5,+5,+8,+9,+5],
+++++++[1,+1,+0,+0,+0,+0,+0]],+dtype=object)|4475542|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This could be one approach -

<pre><code>def numpy_fillna(data):
 # Get lengths of each row of data
 lens = np.array([len(i) for i in data])

 # Mask of valid places in each row
 mask = np.arange(lens.max()) &lt; lens[:,None]

 # Setup output array and put elements from data into masked positions
 out = np.zeros(mask.shape, dtype=data.dtype)
 out[mask] = np.concatenate(data)
 return out
</code></pre>

Sample input, output -

<pre><code>In [222]: # Input object dtype array
 ...: data = np.array([[1, 2, 3, 4],
 ...: [2, 3, 1],
 ...: [5, 5, 5, 5, 8 ,9 ,5],
 ...: [1, 1]])

In [223]: numpy_fillna(data)
Out[223]: 
array([[1, 2, 3, 4, 0, 0, 0],
 [2, 3, 1, 0, 0, 0, 0],
 [5, 5, 5, 5, 8, 9, 5],
 [1, 1, 0, 0, 0, 0, 0]], dtype=object)
</code></pre>

blocks|key|4476375|text|您可以使用pandas而不是numpy：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4476376|In+[1]:+import+pandas+as+pd

In+[2]:+df+=+pd.DataFrame([[1,+2,+3,+4],
+++...:++++++++++++++++++++[2,+3,+1],
+++...:++++++++++++++++++++[5,+5,+5,+5],
+++...:++++++++++++++++++++[1,+1]],+dtype=float)


In+[3]:+df.fillna(0.0).values
Out[3]:+
array([[+1.,++2.,++3.,++4.],
+++++++[+2.,++3.,++1.,++0.],
+++++++[+5.,++5.,++5.,++5.],
+++++++[+1.,++1.,++0.,++0.]])|code-block|syntax|javascript|4476377|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/^0|5|6|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

You could use <a href="http://pandas.pydata.org/">pandas</a> instead of numpy:

<pre><code>In [1]: import pandas as pd

In [2]: df = pd.DataFrame([[1, 2, 3, 4],
 ...: [2, 3, 1],
 ...: [5, 5, 5, 5],
 ...: [1, 1]], dtype=float)


In [3]: df.fillna(0.0).values
Out[3]: 
array([[ 1., 2., 3., 4.],
 [ 2., 3., 1., 0.],
 [ 5., 5., 5., 5.],
 [ 1., 1., 0., 0.]])
</code></pre>

blocks|key|4476400|text|使用np.pad()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4476401|In+[62]:+arr
Out[62]:+
[array([0]),
+array([83,+74]),
+array([87,+61,+23]),
+array([71,++3,+81,+77]),
+array([20,+44,+20,+53,+60]),
+array([54,+36,+74,+35,+49,+54]),
+array([11,+36,++0,+98,+29,+87,+21]),
+array([+1,+22,+62,+51,+45,+40,+36,+86]),
+array([+7,+22,+83,+58,+43,+59,+45,+81,+92]),
+array([68,+78,+70,+67,+77,+64,+58,+88,+13,+56])]

In+[63]:+max_len+=+np.max([len(a)+for+a+in+arr])

In+[64]:+np.asarray([np.pad(a,+(0,+max_len+-+len(a)),+'constant',+constant_values=0)+for+a+in+arr])
Out[64]:+
array([[+0,++0,++0,++0,++0,++0,++0,++0,++0,++0],
+++++++[83,+74,++0,++0,++0,++0,++0,++0,++0,++0],
+++++++[87,+61,+23,++0,++0,++0,++0,++0,++0,++0],
+++++++[71,++3,+81,+77,++0,++0,++0,++0,++0,++0],
+++++++[20,+44,+20,+53,+60,++0,++0,++0,++0,++0],
+++++++[54,+36,+74,+35,+49,+54,++0,++0,++0,++0],
+++++++[11,+36,++0,+98,+29,+87,+21,++0,++0,++0],
+++++++[+1,+22,+62,+51,+45,+40,+36,+86,++0,++0],
+++++++[+7,+22,+83,+58,+43,+59,+45,+81,+92,++0],
+++++++[68,+78,+70,+67,+77,+64,+58,+88,+13,+56]])|code-block|syntax|javascript|4476402|entityMap^0|2|8|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|$]]

use <code>np.pad()</code>.

<pre><code>In [62]: arr
Out[62]: 
[array([0]),
 array([83, 74]),
 array([87, 61, 23]),
 array([71, 3, 81, 77]),
 array([20, 44, 20, 53, 60]),
 array([54, 36, 74, 35, 49, 54]),
 array([11, 36, 0, 98, 29, 87, 21]),
 array([ 1, 22, 62, 51, 45, 40, 36, 86]),
 array([ 7, 22, 83, 58, 43, 59, 45, 81, 92]),
 array([68, 78, 70, 67, 77, 64, 58, 88, 13, 56])]

In [63]: max_len = np.max([len(a) for a in arr])

In [64]: np.asarray([np.pad(a, (0, max_len - len(a)), 'constant', constant_values=0) for a in arr])
Out[64]: 
array([[ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [83, 74, 0, 0, 0, 0, 0, 0, 0, 0],
 [87, 61, 23, 0, 0, 0, 0, 0, 0, 0],
 [71, 3, 81, 77, 0, 0, 0, 0, 0, 0],
 [20, 44, 20, 53, 60, 0, 0, 0, 0, 0],
 [54, 36, 74, 35, 49, 54, 0, 0, 0, 0],
 [11, 36, 0, 98, 29, 87, 21, 0, 0, 0],
 [ 1, 22, 62, 51, 45, 40, 36, 86, 0, 0],
 [ 7, 22, 83, 58, 43, 59, 45, 81, 92, 0],
 [68, 78, 70, 67, 77, 64, 58, 88, 13, 56]])
</code></pre>

blocks|key|4475630|text|data+=+np.array([[1,+2,+3,+4],
+++++++++++++++++[2,+3,+1],
+++++++++++++++++[5,+5,+5,+5],
+++++++++++++++++[1,+1]])
max_len=max([len(i)+for+i+in+data])
np.array([+np.pad(data[i],
+++++++++++(0,max_len-len(data[i])),
+++++++++++'constant',
++++++++++++constant_values=0)+for+i+in+range(len(data))])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4475631|计算各个数组的长度，然后将这些长度中的最大值存储在一个变量中。在此之后，矩阵的所有单独行都在右侧填充0，以匹配最大长度。|unstyled|4475632|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>data = np.array([[1, 2, 3, 4],
 [2, 3, 1],
 [5, 5, 5, 5],
 [1, 1]])
max_len=max([len(i) for i in data])
np.array([ np.pad(data[i],
 (0,max_len-len(data[i])),
 'constant',
 constant_values=0) for i in range(len(data))])
</code></pre>
The lengths of the individual arrays are computed, then the maximum among these lengths is stored in a variable.
After which all the individual rows of the matrix is padded with 0s on the right to match the maximum length.

The functionality I am looking for looks something like this:

<pre><code>data = np.array([[1, 2, 3, 4],
 [2, 3, 1],
 [5, 5, 5, 5],
 [1, 1]])

result = fix(data)
print result

[[ 1. 2. 3. 4.]
 [ 2. 3. 1. 0.]
 [ 5. 5. 5. 5.]
 [ 1. 1. 0. 0.]]
</code></pre>

These data arrays I'm working with are really large so I would really appreciate the most efficient solution.

Edit: Data is read in from disk as a python list of lists.

Numpy: Fix array with rows of different lengths by filling the empty elements with zeros

我正在寻找的功能看起来像这样：data = np.array([[1, 2, 3, 4],                 [2, 3, 1],                 [5, 5, 5, 5],                 [1, 1]])result = fix(data)print result[[ ...

问Numpy:通过用零填充空元素来修复长度不同的数组
EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Numpy:通过用零填充空元素来修复长度不同的数组EN

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Numpy:通过用零填充空元素来修复长度不同的数组
EN