如何在python

Question

我在c ++中有一个for，我想在python中编写它，但我不知道如何处理for中的条件。任何人都可以帮助我吗？

for (int b = (i - a) / 2; a + b <= i; b++); 

Answer 1

直接回答您的问题：

for b in range((i-a)/2,a+b-1):
    print (b)

如果由于某种原因，您需要终止值：

terminatingValue = 0
for b in range((i-a)/2,a+b-1):
    terminatingValue = b
    print (b)

Answer 2

使用while

a=something
i=something
b=(i-a)/2
while(a+b<=i):
    ###do whatever you want in the loop###
    b+=1

Answer 3

您可以使用for循环和 /** * Builds treeChilds from the treeRoots * @param navigationTree The whole navigation Tree containing the treeRoots and their treeChilds * @param navigableRoutes The remaining navigableRoutes that are to be linked to their respectives treeRoots * @returns {RouteDisplay[]} The remaining navigableRoutes after some childs have been taken care of * * @todo : Might need to update the way the childs are filtered : // .filter(routes => !routes.loadChildren && routes.path.includes(child.path)) * @see Doesn't manage duplicate sub route paths (yet) : see @todo */ buildTreeChilds(navigationTree: RouteTreeDisplay[], navigableRoutes: RouteDisplay[]): RouteDisplay[] { // Build treeChilds navigationTree.forEach((treeRoot: RouteTreeDisplay) => { let treeChilds: RouteTreeDisplay[] = []; if (!this.isFirstLevelDone) { treeChilds = navigableRoutes .filter(routes => !routes.loadChildren && routes.parent === treeRoot.path) .map(element => { // Reduce the properties to display return { englishName: element.englishName, frenchName: element.frenchName, materialIcon: element.materialIcon, path: element.path } as RouteTreeDisplay; }); if (treeChilds.length > 0) { treeRoot.children = treeRoot.children.concat(treeChilds); // Clear treeChilds from the navigableRoutes navigableRoutes = navigableRoutes.filter(baseRoute => treeChilds.find(treeChild => treeChild.path === baseRoute.path) ? false : true); treeChilds = []; } } else { // Improve this later on when needed treeRoot.children.forEach(child => { treeChilds = treeChilds.concat( navigableRoutes .filter(routes => !routes.loadChildren && routes.path.includes(child.path)) .map(element => { // Reduce the properties to display return { englishName: element.englishName, frenchName: element.frenchName, materialIcon: element.materialIcon, path: element.path } as RouteTreeDisplay; }) ); if (treeChilds.length > 0) { treeRoot.children = treeRoot.children.concat(treeChilds); // Clear treeChilds from the navigableRoutes navigableRoutes = navigableRoutes.filter(baseRoute => treeChilds.find(treeChild => treeChild.path === baseRoute.path) ? false : true); treeChilds = []; } }); } }); this.isFirstLevelDone = true; return navigableRoutes; } 函数。例如：

range()

但是，由于您的代码，我建议使用while循环：

for i in range(15):
    #code to execute 15 times

Answer 4

你的困惑源于 C 的迭代结构的过于笼统。 Python坚持认为所有循环参数都隐式引用循环索引。对于终止条件，用 b 和没有平等来解决你的不平等： b＆lt;我 - a + 1

现在我们得到 for 循环

for b in range((i-a)/2, (i-a)+1):

一种解释是你的范围从0到i-a，你想要处理它的上半部分。

Answer 5

终止条件a + b <= i相当于b <= i - a，后者（因为Python for使用少于或不等于）b < i - a + 1。

这意味着在Python中你可以写：

for b in range((i - a) / 2, i - a + 1):
    ...

但是从C / C ++转换到Python时，你经常不想像写一样。您可能最好完全重构代码。