是否有可能从通量中获得第一个元素而不将通量转换为流？
答案 0 :(得分：2)
Flux.range(1,10).take(1)应该可以解决问题（此处的range(...)部分只是发出一些样本值；如果您对流进行blockFirst()或subscribe()，则应该看到< strong>“ 1” ）
答案 1 :(得分：1)
除了take(1)之外，如果您需要代表Mono<T>的第一个元素的Flux<T>，则可以使用.next()。
或者，如果您需要第i个元素，请使用.elementAt(i)（但必须确保存在这样的元素，与take和next不同的是，import multiprocessing
print(f'num cpus {multiprocessing.cpu_count():d}')
import sys; print(f'Python {sys.version} on {sys.platform}')
def _process(m):
print(m) #; return m
raise ValueError(m)
args_list = [[i] for i in range(1, 20)]
if __name__ == '__main__':
with multiprocessing.Pool(2) as p:
print([r for r in p.starmap(_process, args_list)])
和num cpus 8
Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 03:13:28)
[Clang 6.0 (clang-600.0.57)] on darwin
1
7
4
10
13
16
19
multiprocessing.pool.RemoteTraceback:
"""
Traceback (most recent call last):
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/multiprocessing/pool.py", line 121, in worker
result = (True, func(*args, **kwds))
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/multiprocessing/pool.py", line 47, in starmapstar
return list(itertools.starmap(args[0], args[1]))
File "/Users/ubik-mac13/Library/Preferences/PyCharm2018.3/scratches/multiprocess_error.py", line 8, in _process
raise ValueError(m)
ValueError: 1
"""
The above exception was the direct cause of the following exception:
Traceback (most recent call last):
File "/Users/ubik-mac13/Library/Preferences/PyCharm2018.3/scratches/multiprocess_error.py", line 18, in <module>
print([r for r in p.starmap(_process, args_list)])
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/multiprocessing/pool.py", line 298, in starmap
return self._map_async(func, iterable, starmapstar, chunksize).get()
File "/Library/Frameworks/Python.framework/Versions/3.7/lib/python3.7/multiprocessing/pool.py", line 683, in get
raise self._value
ValueError: 1
Process finished with exit code 1
仅返回一个空的发布者足够的元素）。