array.array 与 numpy.array 的比较
如果你在Python中创建一个一维数组，使用NumPy这个库有什么好处吗？
3 个回答
0
在存储方面，numpy数组和array.array是可以比较的。下面是一个代码示例，用来比较4字节无符号整数的存储大小。你也可以用其他数据类型来做比较。为了更全面，列表和元组的数据也被加入了比较。
import sys
import numpy as np
from array import array
def getsizeof_deep(obj, seen=None):
"""Recursively finds size of objects"""
size = sys.getsizeof(obj)
if seen is None:
seen = set()
obj_id = id(obj)
if obj_id in seen:
return 0
# Important mark as seen *before* entering recursion to gracefully handle
# self-referential objects
seen.add(obj_id)
if isinstance(obj, dict):
size += sum([getsizeof_deep(v, seen) for v in obj.values()])
size += sum([getsizeof_deep(k, seen) for k in obj.keys()])
elif hasattr(obj, '__dict__'):
size += getsizeof_deep(obj.__dict__, seen)
elif hasattr(obj, '__iter__') and not isinstance(obj, (str, bytes, bytearray)):
size += sum([getsizeof_deep(i, seen) for i in obj])
return size
print("size per element for list, tuple, numpy array, array.array:===============")
for i in range(1, 100, 5):
aa = list(range(i))
n = len(aa)
list_size = getsizeof_deep(aa)
tup_aa = tuple(aa)
tup_size = getsizeof_deep(tup_aa)
nparr = np.array(aa, dtype='uint32')
np_size = getsizeof_deep(nparr)
arr = array('I', aa)#4 byte unsigned integer(in ubuntu)
arr_size = getsizeof_deep(arr)
print('number of element:%s, list %.2f, tuple %.2f, np.array %.2f, arr.array %.2f' % \
(len(aa), list_size/n, tup_size/n, np_size/n, arr_size/n))
在我的机器上，这段代码的输出结果是：
size per element for list, tuple, numpy array, array.array:===============
number of element:1, list 88.00, tuple 72.00, np.array 136.00, arr.array 92.00
number of element:6, list 44.67, tuple 42.00, np.array 49.33, arr.array 42.00
number of element:11, list 40.73, tuple 39.27, np.array 41.45, arr.array 37.45
number of element:16, list 39.25, tuple 38.25, np.array 38.50, arr.array 35.75
number of element:21, list 38.48, tuple 37.71, np.array 36.95, arr.array 34.86
number of element:26, list 38.00, tuple 37.38, np.array 36.00, arr.array 34.31
number of element:31, list 37.68, tuple 37.16, np.array 35.35, arr.array 33.94
number of element:36, list 37.44, tuple 37.00, np.array 34.89, arr.array 33.67
number of element:41, list 37.27, tuple 36.88, np.array 34.54, arr.array 33.46
number of element:46, list 37.13, tuple 36.78, np.array 34.26, arr.array 33.30
number of element:51, list 37.02, tuple 36.71, np.array 34.04, arr.array 33.18
number of element:56, list 36.93, tuple 36.64, np.array 33.86, arr.array 33.07
number of element:61, list 36.85, tuple 36.59, np.array 33.70, arr.array 32.98
number of element:66, list 36.79, tuple 36.55, np.array 33.58, arr.array 32.91
number of element:71, list 36.73, tuple 36.51, np.array 33.46, arr.array 32.85
number of element:76, list 36.68, tuple 36.47, np.array 33.37, arr.array 32.79
number of element:81, list 36.64, tuple 36.44, np.array 33.28, arr.array 32.74
number of element:86, list 36.60, tuple 36.42, np.array 33.21, arr.array 32.70
number of element:91, list 36.57, tuple 36.40, np.array 33.14, arr.array 32.66
number of element:96, list 36.54, tuple 36.38, np.array 33.08, arr.array 32.62
5
这里有个小小的介绍，希望对有需要的人有帮助（参考了@dF.的精彩回答）：
import numpy as np
from array import array
# Fixed size numpy array
def np_fixed(n):
q = np.empty(n)
for i in range(n):
q[i] = i
return q
# Resize with np.resize
def np_class_resize(isize, n):
q = np.empty(isize)
for i in range(n):
if i>=q.shape[0]:
q = np.resize(q, q.shape[0]*2)
q[i] = i
return q
# Resize with the numpy.array method
def np_method_resize(isize, n):
q = np.empty(isize)
for i in range(n):
if i>=q.shape[0]:
q.resize(q.shape[0]*2)
q[i] = i
return q
# Array.array append
def arr(n):
q = array('d')
for i in range(n):
q.append(i)
return q
isize = 1000
n = 10000000
输出结果是：
%timeit -r 10 a = np_fixed(n)
%timeit -r 10 a = np_class_resize(isize, n)
%timeit -r 10 a = np_method_resize(isize, n)
%timeit -r 10 a = arr(n)
1 loop, best of 10: 868 ms per loop
1 loop, best of 10: 2.03 s per loop
1 loop, best of 10: 2.02 s per loop
1 loop, best of 10: 1.89 s per loop
看起来，array.array稍微快一点，而且它的'api'（应用程序接口）能让你少一些麻烦。不过，如果你需要的不仅仅是存储双精度浮点数，那么numpy.resize其实也是个不错的选择（前提是用得当）。