在numpy中安排数据时遇到问题
示例a有数据范围列表：
numpy.array([1,3,5,4,6])
我有数据：
numpy.array([1,2,6,7,8,9,10,11,12,13,14,15,16,17,18,19])
我需要安排数据
numpy.array([
[1,9999,9999]
[2,9999]
[5,9999]
[10,9999]
[14,19]
])
我认为它与diag /对角线/跟踪功能有点相似.
我通常使用基本的迭代来完成这项工作… numpy有这个功能所以它可以执行得更快吗？
最佳答案
以下是一些排列数据的方法：
原文链接：https://www.f2er.com/python/439123.htmlfrom numpy import arange,array,ones,r_,zeros
from numpy.random import randint
def gen_tst(m,n):
a= randint(1,n,m)
b,c= arange(a.sum()),ones((m,n),dtype= int)* 999
return a,b,c
def basic_1(a,c):
# some assumed basic iteration based
n= 0
for k in xrange(len(a)):
m= a[k]
c[k,:m],n= b[n: n+ m],n+ m
def advanced_1(a,c):
# based on Svens answer
cum_a= r_[0,a.cumsum()]
i= arange(len(a)).repeat(a)
j= arange(cum_a[-1])- cum_a[:-1].repeat(a)
c[i,j]= b
def advanced_2(a,c):
# other loopless version
c[arange(c.shape[1])+ zeros((len(a),1),dtype= int)< a[:,None]]= b
还有一些时间：
In []: m,n= 10,100
In []: a,c= gen_tst(m,n)
In []: 1.* a.sum()/ (m* n)
Out[]: 0.531
In []: %timeit advanced_1(a,c)
10000 loops,best of 3: 99.2 us per loop
In []: %timeit advanced_2(a,best of 3: 68 us per loop
In []: %timeit basic_1(a,best of 3: 47.1 us per loop
In []: m,n= 50,500
In []: a,n)
In []: 1.* a.sum()/ (m* n)
Out[]: 0.455
In []: %timeit advanced_1(a,c)
1000 loops,best of 3: 1.03 ms per loop
In []: %timeit advanced_2(a,best of 3: 1.06 ms per loop
In []: %timeit basic_1(a,best of 3: 227 us per loop
In []: m,n= 250,2500
In []: a,n)
In []: 1.* a.sum()/ (m* n)
Out[]: 0.486
In []: %timeit advanced_1(a,c)
10 loops,best of 3: 30.4 ms per loop
In []: %timeit advanced_2(a,best of 3: 32.4 ms per loop
In []: %timeit basic_1(a,best of 3: 2 ms per loop
所以基本的迭代似乎非常有效.
更新：
当然,基于迭代的基本实现的性能仍然可以进一步提高.作为一个起点建议;例如考虑这个(基于减少加法的基本迭代)：
def basic_2(a,c):
n= 0
for k,m in enumerate(a):
nm= n+ m
c[k,n= b[n: nm],nm