在for循环期间从Python列表中删除内容

这是我的密码：

toBe =[]
#Check if there is any get request
if request.GET.items() != []:
    for x in DB:

        try:
            #This is removing the PMT that is spcific to the name
            if request.GET['pmtName'] != "None":
                if not request.GET['pmtName'] in x['tags']:
                    print x['title'], x['tags']
                    toBe.append(x)
                    continue

            #This is removing the peak stuff
            if int(request.GET['peakMin']):
                if int(request.GET['peakMin']) < int(x['charge_peak']):
                    toBe.append(x)
                    continue
            if int(request.GET['peakMax']):
                if int(request.GET['peakMax']) > int(x['charge_peak']):
                    toBe.append(x)
                    continue
            if int(request.GET['widthMin']):
                if int(request.GET['widthMin']) < int(x['charge_width']):
                    toBe.append(x)
                    continue
            if int(request.GET['widthMax']):
                if int(request.GET['widthMax']) > int(x['charge_width']):
                    toBe.append(x)
                    continue
        except:
            pass
#TODO: Stupid hack, this needs to be fixed
for x in toBe:
    DB.remove(x)
del toBe

toBe=[]
#检查是否有任何get请求
if request.GET.items（）！=[]:
对于以DB表示的x：
尝试：
#这将删除与名称相关的PMT
如果请求，则获取['pmtName']！=“无”：
如果不是请求，则在x['tags']中获取['pmtName']：
打印x['title'，x['tags']
toBe.append（x）
持续
#这是删除峰值的东西
if int（request.GET['peakMin']）：
如果int（request.GET['peakMin']）int（x['charge\u peak']）：
toBe.append（x）
持续
if int（request.GET['widthMin']）：
如果int（request.GET['widthMin']）int（x['charge\u width']）：
toBe.append（x）
持续
除：
通过
#TODO:愚蠢的黑客，这需要修复
对于x in toBe：
DB.remove（x）
德尔托比

基本上，我想删除该项，然后跳到下一项。问题是，当这种情况发生时，它会打乱列表的顺序，并跳过一些。有人知道这方面的工作吗？或者只是一种不同的方式

---

感谢

在DB[：]中输入x:

制作列表的副本DB，这样您可以在修改原始列表的同时对其进行迭代。小心——记忆密集且缓慢

更好的方法是在列表上创建另一层，只生成部分值，然后在以后需要时迭代该层。您可以使用发电机执行此操作：

def db_view( DB ):
    for x in DB:

        #This is removing the PMT that is spcific to the name
        if request.GET.get( 'pmtName', None ) not in x['tags']:
                print x['title'], x['tags']
                continue

        #This is removing the peak stuff
        if int(request.GET['peakMin']):
            if int(request.GET['peakMin']) < int(x['charge_peak']):
                continue

        if int(request.GET['peakMax']):
            if int(request.GET['peakMax']) > int(x['charge_peak']):
                continue

        if int(request.GET['widthMin']):
            if int(request.GET['widthMin']) < int(x['charge_width']):
                continue

        if int(request.GET['widthMax']):
            if int(request.GET['widthMax']) > int(x['charge_width']):
                continue

        yield x

---

我通常看到的关于这类问题的答案是，如果要向后循环列表的话。以下是我在我的一个程序中如何做到这一点：

for i in range(len(my_list)-1,-1,-1):
    # do something

---

即使我将项目添加到列表中，这也有效。他们说你可以用“for i in list[：：-1]：”来代替。我没有尝试过这样做。

您正在运行相同的

请求插值。为
x的每个值获取
。相反，您可以一次性构建一个可重用的过滤函数列表

例如：

if request.GET:
    filters = []
    if 'pmtName' in request.GET:
        n = request.GET['pmtName']
        filters.append(lambda x: n not in x['tags'])
    if 'peakMin' in request.GET and request.GET['peakMin'].isdigit():
        n = int(request.GET['peakMin'])
        filters.append(lambda x: n < int(x['charge_peak']))
    if 'peakMax' in request.GET and request.GET['peakMax'].isdigit():
        n = int(request.GET['peakMax'])
        filters.append(lambda x: n > int(x['charge_peak']))
    if 'widthMin' in request.GET and request.GET['widthMin'].isdigit():
        n = int(request.GET['widthMin'])
        filters.append(lambda x: n < int(x['charge_width']))
    if 'widthMax' in request.GET and request.GET['widthMax'].isdigit():
        n = int(request.GET['widthMax'])
        filters.append(lambda x: n > int(x['charge_width']))

或者创建一个生成器，该生成器将返回所有过滤器失效时的DB值：

filtered_DB = ( x for x in DB if all(not f(x) for f in filters))

remove_these = [ x for x in DB if any(f(x) for f in filters)]
for item in remove_these:
    DB.remove(item)

filtered_DB = ( x for x in DB if all(not f(x) for f in filters))