我有以下情况：
for x1 in range(x1, x2):
for x2 in range(x3, x4):
for x3 ...
...
f(x1, x2, x3, ...)
如何将此转换为一种机制，我只告诉python使 n 嵌套循环，其中变量名是x1，x2，x3，x4，...？我当然不想手动编写所有可能性，因为可能存在很多维度。
答案 0 :(得分：6)
您要做的是迭代产品。使用itertools.product。
import itertools
ranges = [range(x1, x2), range(x3, x4), ...]
for xs in itertools.product(*ranges):
f(*xs)
import itertools
ranges = [range(0, 2), range(1, 3), range(2, 4)]
for xs in itertools.product(*ranges):
print(*xs)
0 1 2
0 1 3
0 2 2
0 2 3
1 1 2
1 1 3
1 2 2
1 2 3
答案 1 :(得分：4)
itertools是一个很棒的包：
from itertools import product
x1 = 3; x2 = 4
x3 = 0; x4 = 2
x5 = 42; x6 = 42
for x, y, z in product(range(x1, x2), range(x3, x4), range(x4, x5)):
print(x, y, z)
给出
3 0 2
3 0 3
3 0 4
3 0 5
3 0 6
3 0 7
3 0 8
3 0 9
3 0 10
3 0 11
3 0 12
3 0 13
3 0 14
3 0 15
3 0 16
3 0 17
3 0 18
3 0 19
3 0 20
3 0 21
3 0 22
3 0 23
3 0 24
3 0 25
3 0 26
3 0 27
3 0 28
3 0 29
3 0 30
3 0 31
3 0 32
3 0 33
3 0 34
3 0 35
3 0 36
3 0 37
3 0 38
3 0 39
3 0 40
3 0 41
3 1 2
3 1 3
3 1 4
3 1 5
3 1 6
3 1 7
3 1 8
3 1 9
3 1 10
3 1 11
3 1 12
3 1 13
3 1 14
3 1 15
3 1 16
3 1 17
3 1 18
3 1 19
3 1 20
3 1 21
3 1 22
3 1 23
3 1 24
3 1 25
3 1 26
3 1 27
3 1 28
3 1 29
3 1 30
3 1 31
3 1 32
3 1 33
3 1 34
3 1 35
3 1 36
3 1 37
3 1 38
3 1 39
3 1 40
3 1 41
def product(*args):
pools = map(tuple, args)
result = [[]]
for pool in pools:
result = [x + [y] for x in result for y in pool]
for prod in result:
yield tuple(prod)
答案 2 :(得分：0)
您还可以使用numpy.ndindex来实现您的目标：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="ie=edge">
<link rel="stylesheet" href="../css/style.css" />
<title>Floza Web Design | Conception web</title>
</head>
<body>
<header>
<div class="nav" id="nav">
<a href="javascript:void(0)" id="close" class="close" onclick="closeNav()">×</a>
<a href="#">Accueil</a>
<a href="#">Réalisations</a>
<a href="#">Notre équipe</a>
<a href="#">Nos services</a>
<a href="#">Contact</a>
</div>
<span id="open" class="open" onclick="openNav()">☰</span>
<script type="text/javascript">
function openNav(){
document.getElementById('open').style.display = "none";
document.getElementById('nav').style.width = "100%";
}
function closeNav(){
document.getElementById('nav').style.width = "0";
document.getElementById('open').style.display = "block";
}
</script>
</header>
<div class="div"></div>
</body>
</html>当你已经在脚本中使用numpy之类的东西时，这特别有用。您必须将'起点'（即x1，x2，...）添加到每个维度（如果它不是0）。
答案 3 :(得分：-1)
使用多处理！
import multiprocessing
from itertools import product
results = []
def callback(return_value):
# this stores results
results.append(return_value)
if __name__=="__main__" :
pool = multiprocessing.Pool(4)
args = product(range1, range2, range)
for x, y, z in args:
pool.apply_async(f,(x,y,z),call_back=callback)
pool.close()
pool.join()
现在您可以在充分利用机器的全部容量的同时运行您的功能！