给定输入
A = np.array([[1,2,3],[4,5,6],[7,8,9]])
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
需要输出:
array([[2, 3],
[4, 6],
[7, 8]])
使用迭代或循环很容易做到这一点,但应该有一个简洁的方法来做到这一点,而不使用循环。感谢
答案 0 :(得分:7)
方法#1
使用masking
-
A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)
示例运行 -
In [395]: A
Out[395]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
In [396]: A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)
Out[396]:
array([[2, 3],
[4, 6],
[7, 8]])
方法#2
使用非对角线元素的常规模式,可以使用范围数组广播添加进行跟踪 -
m = A.shape[0]
idx = (np.arange(1,m+1) + (m+1)*np.arange(m-1)[:,None]).reshape(m,-1)
out = A.ravel()[idx]
方法#3 (Strides Strikes!)
滥用先前方法中的非对角元素的常规模式,我们可以引入np.lib.stride_tricks.as_strided
和一些slicing
帮助,如此 -
m = A.shape[0]
strided = np.lib.stride_tricks.as_strided
s0,s1 = A.strides
out = strided(A.ravel()[1:], shape=(m-1,m), strides=(s0+s1,s1)).reshape(m,-1)
运行时测试
作为funcs的方法:
def skip_diag_masking(A):
return A[~np.eye(A.shape[0],dtype=bool)].reshape(A.shape[0],-1)
def skip_diag_broadcasting(A):
m = A.shape[0]
idx = (np.arange(1,m+1) + (m+1)*np.arange(m-1)[:,None]).reshape(m,-1)
return A.ravel()[idx]
def skip_diag_strided(A):
m = A.shape[0]
strided = np.lib.stride_tricks.as_strided
s0,s1 = A.strides
return strided(A.ravel()[1:], shape=(m-1,m), strides=(s0+s1,s1)).reshape(m,-1)
计时 -
In [528]: A = np.random.randint(11,99,(5000,5000))
In [529]: %timeit skip_diag_masking(A)
...: %timeit skip_diag_broadcasting(A)
...: %timeit skip_diag_strided(A)
...:
10 loops, best of 3: 56.1 ms per loop
10 loops, best of 3: 82.1 ms per loop
10 loops, best of 3: 32.6 ms per loop
答案 1 :(得分:1)
只需要numpy,假设一个方阵:
new_A = numpy.delete(A,range(0,A.shape[0]**2,(A.shape[0]+1))).reshape(A.shape[0],(A.shape[1]-1))
答案 2 :(得分:1)
我知道我参加这个聚会很晚,但是我相信这是一个简单的解决方案。所以要删除对角线吗?好的,很酷:
`
arr = np.array([[1,2,3],[4,5,6],[7,8,9]]).astype(np.float)
np.fill_diagonal(arr, np.nan)
arr[~np.isnan(arr)].reshape(arr.shape[0], arr.shape[1] - 1)
答案 3 :(得分:1)
解决步骤:
range(0, len(x_no_diag), len(x) + 1)
功能:
import numpy as np
def remove_diag(x):
x_no_diag = np.ndarray.flatten(x)
x_no_diag = np.delete(x_no_diag, range(0, len(x_no_diag), len(x) + 1), 0)
x_no_diag = x_no_diag.reshape(len(x), len(x) - 1)
return x_no_diag
示例:
>>> x = np.random.randint(5, size=(3,3))
array([[0, 2, 3],
[3, 4, 1],
[2, 4, 0]])
>>> remove_diag(x)
array([[2, 3],
[3, 1],
[2, 4]])
答案 4 :(得分:0)
如果您不介意创建新数组,则可以使用列表推导。
A = np.array([A[i][A[i] != A[i][i]] for i in range(len(A))])
运行与@Divakar相同的方法,
A = np.random.randint(11,99,(5000,5000))
skip_diag_masking
85.7 ms ± 1.55 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_broadcasting
163 ms ± 1.77 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_strided
52.5 ms ± 2.54 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
skip_diag_list_comp
101 ms ± 347 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
答案 5 :(得分:0)
也许基于Divakar's first solution但使用import pandas as pd
df = pd.DataFrame({'Date': {0: '2020-08-29',
1: '2020-08-31',
2: '2020-09-01',
3: '2020-09-25',
4: '2020-09-26',
5: '2020-09-30'},
'value': {0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0}})
df['Date'] = pd.to_datetime(df['Date']).sort_values()
m1 = df.groupby(df['Date'].dt.month)['Date'].transform('max')
m2 = df['Date'].shift(-1)
m = (m1 == m2)
df['Max Date Per Month'] = m1
df['Shifted Date'] = m2
df['Keep/Drop'] = m
df
Out[33]:
Date value Max Date Per Month Shifted Date Keep/Drop
0 2020-08-29 0 2020-08-31 2020-08-31 True
1 2020-08-31 0 2020-08-31 2020-09-01 False
2 2020-09-01 0 2020-09-30 2020-09-25 False
3 2020-09-25 0 2020-09-30 2020-09-26 False
4 2020-09-26 0 2020-09-30 2020-09-30 True
5 2020-09-30 0 2020-09-30 NaT False
而非len(array)
的最干净的方法是:
array.shape[0]
答案 6 :(得分:-1)
另一种方法是使用numpy.delete()。假设方阵,您可以使用:
numpy.delete(A,range(0,A.shape[0]**2,A.shape[0])).reshape(A.shape[0],A.shape[1]-1)