Python 是否有按功能划分的任何 numpy 组？
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Is there any numpy group by function?
提问by John Dow
Is there any function in numpy to group this array down below by the first column?
numpy 中是否有任何函数可以将这个数组按第一列分组？
I couldn't find any good answer over the internet..
我在互联网上找不到任何好的答案..
>>> a
array([[ 1, 275],
[ 1, 441],
[ 1, 494],
[ 1, 593],
[ 2, 679],
[ 2, 533],
[ 2, 686],
[ 3, 559],
[ 3, 219],
[ 3, 455],
[ 4, 605],
[ 4, 468],
[ 4, 692],
[ 4, 613]])
Wanted output:
想要的输出：
array([[[275, 441, 494, 593]],
[[679, 533, 686]],
[[559, 219, 455]],
[[605, 468, 692, 613]]], dtype=object)
回答by Vincent J
Inspired by Eelco Hoogendoorn's library, but without his library, and using the fact that the first column of your array is always increasing.
受 Eelco Hoogendoorn 库的启发，但没有他的库，并使用数组的第一列始终增加的事实。
>>> np.split(a[:, 1], np.cumsum(np.unique(a[:, 0], return_counts=True)[1])[:-1])
[array([275, 441, 494, 593]),
array([679, 533, 686]),
array([559, 219, 455]),
array([605, 468, 692, 613])]
I didn't "timeit" but this is probably the faster way to achieve the question :
我没有“timeit”，但这可能是解决问题的更快方法：
- No python native loop
- Result lists are numpy arrays, in case you need to make other numpy operations on them, no new conversion will be needed
- Complexity like O(n)
- 没有python本机循环
- 结果列表是 numpy 数组，如果您需要对它们进行其他 numpy 操作，则不需要新的转换
- 复杂度像 O(n)
PS: I wrote a similar line because I needed to "group by" the results of np.nonzero:
PS：我写了一个类似的行，因为我需要对 np.nonzero 的结果进行“分组”：
>>> indexes, values = np.nonzero(...)
>>> np.split(values, np.cumsum(np.unique(indexes, return_counts=True)[1]))
回答by Eelco Hoogendoorn
The numpy_indexedpackage (disclaimer: I am its author) aims to fill this gap in numpy. All operations in numpy-indexed are fully vectorized, and no O(n^2) algorithms were harmed during the making of this library.
该numpy_indexed包（免责声明：我是它的作者）的目标，以填补在numpy的这一空白。numpy-indexed 中的所有操作都是完全矢量化的，并且在该库的制作过程中没有损坏 O(n^2) 算法。
import numpy_indexed as npi
npi.group_by(a[:, 0]).split(a[:, 1])
Note that it is usually more efficient to directly compute relevant properties over such groups (ie, group_by(keys).mean(values)), rather than first splitting into a list / jagged array.
请注意，直接计算此类组的相关属性通常更有效（即 group_by(keys).mean(values)），而不是首先拆分为列表/锯齿状数组。
回答by Piotr
Numpy is not very handy here because the desired output is not an array of integers (it is an array of list objects).
Numpy 在这里不是很方便，因为所需的输出不是整数数组（它是一个列表对象数组）。
I suggest either the pure Python way...
我建议使用纯 Python 方式...
from collections import defaultdict
%%timeit
d = defaultdict(list)
for key, val in a:
d[key].append(val)
10.7 μs ± 156 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
# result:
defaultdict(list,
{1: [275, 441, 494, 593],
2: [679, 533, 686],
3: [559, 219, 455],
4: [605, 468, 692, 613]})
...or the pandas way:
...或熊猫方式：
import pandas as pd
%%timeit
df = pd.DataFrame(a, columns=["key", "val"])
df.groupby("key").val.apply(pd.Series.tolist)
979 μs ± 3.3 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
# result:
key
1 [275, 441, 494, 593]
2 [679, 533, 686]
3 [559, 219, 455]
4 [605, 468, 692, 613]
Name: val, dtype: object
回答by Gioelelm
n = np.unique(a[:,0])
np.array( [ list(a[a[:,0]==i,1]) for i in n] )
outputs:
输出：
array([[275, 441, 494, 593], [679, 533, 686], [559, 219, 455],
[605, 468, 692, 613]], dtype=object)
回答by ns63sr
Simplifying the answer of Vincent Jone can use return_index = Trueinstead of return_counts = Trueand get rid of the cumsum:
简化文森特 J的答案，可以使用return_index = True代替return_counts = True并去掉cumsum：
np.split(a[:,1], np.unique(idx,return_index = True)[1][1:])
Output
输出
[array([275, 441, 494, 593]),
array([679, 533, 686]),
array([559, 219, 455]),
array([605, 468, 692, 613])]
回答by Guido Mocha
given X as array of items you want to be grouped and y (1D array) as corresponding groups, following function does the grouping with numpy:
给定 X 作为要分组的项目数组，将 y （一维数组）作为相应的组，以下函数使用numpy进行分组：
def groupby(X, y):
y = np.asarray(y)
X = np.asarray(X)
y_uniques = np.unique(y)
return [X[y==yi] for yi in y_uniques]
So, groupby(a[:,1], a[:,0])returns
[array([275, 441, 494, 593]), array([679, 533, 686]), array([559, 219, 455]), array([605, 468, 692, 613])]
所以，groupby(a[:,1], a[:,0])返回
[array([275, 441, 494, 593]), array([679, 533, 686]), array([559, 219, 455]), array([605, 468, 692, 613])]
回答by user2251346
I used np.unique() followed by np.extract()
我使用 np.unique() 后跟 np.extract()
unique = np.unique(a[:, 0:1])
answer = []
for element in unique:
present = a[:,0]==element
answer.append(np.extract(present,a[:,-1]))
print (answer)
[array([275, 441, 494, 593]), array([679, 533, 686]), array([559, 219, 455]), array([605, 468, 692, 613])]
[array([275, 441, 494, 593]), array([679, 533, 686]), array([559, 219, 455]), array([605, 468, 692, 613])]
回答by javadba
We might also find it useful to generate a dict:
我们可能还会发现生成一个很有用dict：
def groupby(X):
X = np.asarray(X)
x_uniques = np.unique(X)
return {xi:X[X==xi] for xi in x_uniques}
Let's try it out:
让我们试试看：
X=[1,1,2,2,3,3,3,3,4,5,6,7,7,8,9,9,1,1,1]
groupby(X)
Out[9]:
{1: array([1, 1, 1, 1, 1]),
2: array([2, 2]),
3: array([3, 3, 3, 3]),
4: array([4]),
5: array([5]),
6: array([6]),
7: array([7, 7]),
8: array([8]),
9: array([9, 9])}
Note this by itself is not super compelling - but if we make Xan objector namedtupleand then provide a groupbyfunction it becomes more interesting. Will put that in later.
请注意，这本身并不是非常引人注目 - 但如果我们制作X一个objectornamedtuple然后提供一个groupby函数，它会变得更有趣。以后会放上来。
