Python Numpy：查找范围内元素的索引
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Numpy: find index of the elements within range
提问by Bob
I have a numpy array of numbers, for example,
例如，我有一个 numpy 数字数组，
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
I would like to find all the indexes of the elements within a specific range. For instance, if the range is (6, 10), the answer should be (3, 4, 5). Is there a built-in function to do this?
我想找到特定范围内元素的所有索引。例如，如果范围是 (6, 10)，那么答案应该是 (3, 4, 5)。是否有内置函数来执行此操作？
采纳答案by deinonychusaur
You can use np.whereto get indices and np.logical_andto set two conditions:
您可以使用np.where来获取索引并np.logical_and设置两个条件：
import numpy as np
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.where(np.logical_and(a>=6, a<=10))
# returns (array([3, 4, 5]),)
回答by Bi Rico
I thought I would add this because the ain the example you gave is sorted:
我想我会添加这个，因为a在你给出的例子中是排序的：
import numpy as np
a = [1, 3, 5, 6, 9, 10, 14, 15, 56]
start = np.searchsorted(a, 6, 'left')
end = np.searchsorted(a, 10, 'right')
rng = np.arange(start, end)
rng
# array([3, 4, 5])
回答by tiago
As in @deinonychusaur's reply, but even more compact:
就像@deinonychusaur 的回复一样，但更紧凑：
In [7]: np.where((a >= 6) & (a <=10))
Out[7]: (array([3, 4, 5]),)
回答by Vigneshwaran Narayanan
s=[52, 33, 70, 39, 57, 59, 7, 2, 46, 69, 11, 74, 58, 60, 63, 43, 75, 92, 65, 19, 1, 79, 22, 38, 26, 3, 66, 88, 9, 15, 28, 44, 67, 87, 21, 49, 85, 32, 89, 77, 47, 93, 35, 12, 73, 76, 50, 45, 5, 29, 97, 94, 95, 56, 48, 71, 54, 55, 51, 23, 84, 80, 62, 30, 13, 34]
dic={}
for i in range(0,len(s),10):
dic[i,i+10]=list(filter(lambda x:((x>=i)&(x<i+10)),s))
print(dic)
for keys,values in dic.items():
print(keys)
print(values)
Output:
输出：
(0, 10)
[7, 2, 1, 3, 9, 5]
(20, 30)
[22, 26, 28, 21, 29, 23]
(30, 40)
[33, 39, 38, 32, 35, 30, 34]
(10, 20)
[11, 19, 15, 12, 13]
(40, 50)
[46, 43, 44, 49, 47, 45, 48]
(60, 70)
[69, 60, 63, 65, 66, 67, 62]
(50, 60)
[52, 57, 59, 58, 50, 56, 54, 55, 51]
回答by Abhishek
a = np.array([1,2,3,4,5,6,7,8,9])
b = a[(a>2) & (a<8)]
回答by Deba Pratim Saha
You can use np.clip()to achieve the same:
您可以使用以下np.clip()方法实现相同的目的：
a = [1, 3, 5, 6, 9, 10, 14, 15, 56]
np.clip(a,6,10)
However, it holds the values less than and greater than 6 and 10 respectively.
但是，它分别保存小于和大于 6 和 10 的值。
回答by Abhishek
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.argwhere((a>=6) & (a<=10))
回答by Nathan
This code snippet returns all the numbers in a numpy array between two values:
此代码段返回两个值之间的 numpy 数组中的所有数字：
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56] )
a[(a>6)*(a<10)]
It works as following: (a>6) returns a numpy array with True (1) and False (0), so does (a<10). By multiplying these two together you get an array with either a True, if both statements are True (because 1x1 = 1) or False (because 0x0 = 0 and 1x0 = 0).
它的工作原理如下：(a>6) 返回一个包含 True (1) 和 False (0) 的 numpy 数组，(a<10) 也是如此。通过将这两者相乘，您将得到一个带有 True 的数组，如果两个语句都是 True（因为 1x1 = 1）或 False（因为 0x0 = 0 和 1x0 = 0）。
The part a[...] returns all values of array a where the array between brackets returns a True statement.
a[...] 部分返回数组 a 的所有值，其中括号之间的数组返回 True 语句。
Of course you can make this more complicated by saying for instance
当然，您可以通过说例如使这更复杂
...*(1-a<10)
which is similar to an "and Not" statement.
这类似于“and Not”语句。
回答by cvanelteren
This may not be the prettiest, but works for any dimension
这可能不是最漂亮的，但适用于任何维度
a = np.array([[-1,2], [1,5], [6,7], [5,2], [3,4], [0, 0], [-1,-1]])
ranges = (0,4), (0,4)
def conditionRange(X : np.ndarray, ranges : list) -> np.ndarray:
idx = set()
for column, r in enumerate(ranges):
tmp = np.where(np.logical_and(X[:, column] >= r[0], X[:, column] <= r[1]))[0]
if idx:
idx = idx & set(tmp)
else:
idx = set(tmp)
idx = np.array(list(idx))
return X[idx, :]
b = conditionRange(a, ranges)
print(b)
回答by Jorge Cruz
Other way is with:
另一种方法是：
np.vectorize(lambda x: 6 <= x <= 10)(a)
which returns:
返回：
array([False, False, False, True, True, True, False, False, False])
It is sometimes useful for masking time series, vectors, etc.
它有时可用于屏蔽时间序列、向量等。
