在Python多处理库中，是否有支持多个参数的pool.map变体？
import multiprocessing
text = "test"
def harvester(text, case):
X = case[0]
text + str(X)
if __name__ == '__main__':
pool = multiprocessing.Pool(processes=6)
case = RAW_DATASET
pool.map(harvester(text, case), case, 1)
pool.close()
pool.join()
当前回答
这可能是另一种选择。技巧在于包装器函数，它返回传递给pool.map的另一个函数。下面的代码读取一个输入数组，对于其中的每个（唯一）元素，返回该元素在数组中出现的次数（即计数）。例如，如果输入是
np.eye(3) = [ [1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
然后零出现6次，一出现3次
import numpy as np
from multiprocessing.dummy import Pool as ThreadPool
from multiprocessing import cpu_count
def extract_counts(label_array):
labels = np.unique(label_array)
out = extract_counts_helper([label_array], labels)
return out
def extract_counts_helper(args, labels):
n = max(1, cpu_count() - 1)
pool = ThreadPool(n)
results = {}
pool.map(wrapper(args, results), labels)
pool.close()
pool.join()
return results
def wrapper(argsin, results):
def inner_fun(label):
label_array = argsin[0]
counts = get_label_counts(label_array, label)
results[label] = counts
return inner_fun
def get_label_counts(label_array, label):
return sum(label_array.flatten() == label)
if __name__ == "__main__":
img = np.ones([2,2])
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
img = np.eye(3)
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
img = np.random.randint(5, size=(3, 3))
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
你应该得到：
input array:
[[1. 1.]
[1. 1.]]
label counts: {1.0: 4}
========
input array:
[[1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
label counts: {0.0: 6, 1.0: 3}
========
input array:
[[4 4 0]
[2 4 3]
[2 3 1]]
label counts: {0: 1, 1: 1, 2: 2, 3: 2, 4: 3}
========
其他回答
另一个简单的选择是将函数参数包装在元组中，然后包装应该在元组中传递的参数。在处理大量数据时，这可能并不理想。我相信它会为每个元组创建副本。
from multiprocessing import Pool
def f((a,b,c,d)):
print a,b,c,d
return a + b + c +d
if __name__ == '__main__':
p = Pool(10)
data = [(i+0,i+1,i+2,i+3) for i in xrange(10)]
print(p.map(f, data))
p.close()
p.join()
以某种随机顺序给出输出：
0 1 2 3
1 2 3 4
2 3 4 5
3 4 5 6
4 5 6 7
5 6 7 8
7 8 9 10
6 7 8 9
8 9 10 11
9 10 11 12
[6, 10, 14, 18, 22, 26, 30, 34, 38, 42]
将Python 3.3+与pool.starmap（）一起使用：
from multiprocessing.dummy import Pool as ThreadPool
def write(i, x):
print(i, "---", x)
a = ["1","2","3"]
b = ["4","5","6"]
pool = ThreadPool(2)
pool.starmap(write, zip(a,b))
pool.close()
pool.join()
结果：
1 --- 4
2 --- 5
3 --- 6
如果您喜欢，还可以zip（）更多参数：zip（a，b，c，d，e）
如果希望将常量值作为参数传递：
import itertools
zip(itertools.repeat(constant), a)
如果您的函数应该返回以下内容：
results = pool.starmap(write, zip(a,b))
这将提供一个包含返回值的列表。
如何获取多个参数：
def f1(args):
a, b, c = args[0] , args[1] , args[2]
return a+b+c
if __name__ == "__main__":
import multiprocessing
pool = multiprocessing.Pool(4)
result1 = pool.map(f1, [ [1,2,3] ])
print(result1)
这可能是另一种选择。技巧在于包装器函数，它返回传递给pool.map的另一个函数。下面的代码读取一个输入数组，对于其中的每个（唯一）元素，返回该元素在数组中出现的次数（即计数）。例如，如果输入是
np.eye(3) = [ [1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
然后零出现6次，一出现3次
import numpy as np
from multiprocessing.dummy import Pool as ThreadPool
from multiprocessing import cpu_count
def extract_counts(label_array):
labels = np.unique(label_array)
out = extract_counts_helper([label_array], labels)
return out
def extract_counts_helper(args, labels):
n = max(1, cpu_count() - 1)
pool = ThreadPool(n)
results = {}
pool.map(wrapper(args, results), labels)
pool.close()
pool.join()
return results
def wrapper(argsin, results):
def inner_fun(label):
label_array = argsin[0]
counts = get_label_counts(label_array, label)
results[label] = counts
return inner_fun
def get_label_counts(label_array, label):
return sum(label_array.flatten() == label)
if __name__ == "__main__":
img = np.ones([2,2])
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
img = np.eye(3)
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
img = np.random.randint(5, size=(3, 3))
out = extract_counts(img)
print('input array: \n', img)
print('label counts: ', out)
print("========")
你应该得到：
input array:
[[1. 1.]
[1. 1.]]
label counts: {1.0: 4}
========
input array:
[[1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]
label counts: {0.0: 6, 1.0: 3}
========
input array:
[[4 4 0]
[2 4 3]
[2 3 1]]
label counts: {0: 1, 1: 1, 2: 2, 3: 2, 4: 3}
========
更好的方法是使用修饰符，而不是手工编写包装函数。特别是当您有很多函数要映射时，装饰器将通过避免为每个函数编写包装器来节省时间。通常，修饰函数是不可选择的，但是我们可以使用functools来解决它。更多讨论可以在这里找到。
以下是示例：
def unpack_args(func):
from functools import wraps
@wraps(func)
def wrapper(args):
if isinstance(args, dict):
return func(**args)
else:
return func(*args)
return wrapper
@unpack_args
def func(x, y):
return x + y
然后你可以用压缩的参数来映射它：
np, xlist, ylist = 2, range(10), range(10)
pool = Pool(np)
res = pool.map(func, zip(xlist, ylist))
pool.close()
pool.join()
当然，您可能总是在Python3中使用Pool.starmap（>=3.3），正如其他答案中提到的那样。
