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Python：在for循环中“跳出”if语句

8 投票
5 回答
47591 浏览
提问于 2025-04-16 23:22
X 0 randomValue 0.0775612781213 prob container 0.0294117647059
X 1 randomValue 0.0775612781213 prob container 0.0294117647059
X 2 randomValue 0.0775612781213 prob container 0.0294117647059
X 3 randomValue 0.0775612781213 prob container 0.0294117647059
O 4 randomValue 0.0775612781213 prob container 0.147058823529
O 5 randomValue 0.0775612781213 prob container 0.235294117647
O 6 randomValue 0.0775612781213 prob container 0.441176470588
O 7 randomValue 0.0775612781213 prob container 0.588235294118
O 8 randomValue 0.0775612781213 prob container 0.676470588235
O 9 randomValue 0.0775612781213 prob container 0.764705882353
O 10 randomValue 0.0775612781213 prob container 0.794117647059
O 11 randomValue 0.0775612781213 prob container 0.823529411765
O 12 randomValue 0.0775612781213 prob container 0.823529411765
O 13 randomValue 0.0775612781213 prob container 0.852941176471
O 14 randomValue 0.0775612781213 prob container 0.882352941176
O 15 randomValue 0.0775612781213 prob container 0.882352941176
O 16 randomValue 0.0775612781213 prob container 0.911764705882
O 17 randomValue 0.0775612781213 prob container 0.911764705882
O 18 randomValue 0.0775612781213 prob container 0.911764705882
O 19 randomValue 0.0775612781213 prob container 0.911764705882
O 20 randomValue 0.0775612781213 prob container 0.911764705882
O 21 randomValue 0.0775612781213 prob container 0.941176470588
O 22 randomValue 0.0775612781213 prob container 0.941176470588
O 23 randomValue 0.0775612781213 prob container 0.970588235294
O 24 randomValue 0.0775612781213 prob container 0.970588235294
O 25 randomValue 0.0775612781213 prob container 0.970588235294
O 26 randomValue 0.0775612781213 prob container 0.970588235294
O 27 randomValue 0.0775612781213 prob container 0.970588235294
O 28 randomValue 0.0775612781213 prob container 1.0
X 0 randomValue 0.803308376497 prob container 0.0294117647059
X 1 randomValue 0.803308376497 prob container 0.0294117647059
X 2 randomValue 0.803308376497 prob container 0.0294117647059
X 3 randomValue 0.803308376497 prob container 0.0294117647059
X 4 randomValue 0.803308376497 prob container 0.0294117647059
X 5 randomValue 0.803308376497 prob container 0.0294117647059
X 6 randomValue 0.803308376497 prob container 0.0882352941176
X 7 randomValue 0.803308376497 prob container 0.0882352941176
X 8 randomValue 0.803308376497 prob container 0.0882352941176
X 9 randomValue 0.803308376497 prob container 0.117647058824
X 10 randomValue 0.803308376497 prob container 0.147058823529
X 11 randomValue 0.803308376497 prob container 0.205882352941
X 12 randomValue 0.803308376497 prob container 0.264705882353
X 13 randomValue 0.803308376497 prob container 0.294117647059
X 14 randomValue 0.803308376497 prob container 0.382352941176
X 15 randomValue 0.803308376497 prob container 0.441176470588
X 16 randomValue 0.803308376497 prob container 0.470588235294
X 17 randomValue 0.803308376497 prob container 0.470588235294
X 18 randomValue 0.803308376497 prob container 0.529411764706
X 19 randomValue 0.803308376497 prob container 0.588235294118
X 20 randomValue 0.803308376497 prob container 0.647058823529
X 21 randomValue 0.803308376497 prob container 0.764705882353
O 22 randomValue 0.803308376497 prob container 0.823529411765
O 23 randomValue 0.803308376497 prob container 0.882352941176
O 24 randomValue 0.803308376497 prob container 0.970588235294
O 25 randomValue 0.803308376497 prob container 0.970588235294
O 26 randomValue 0.803308376497 prob container 1.0

我明白一个if语句是不能被“中断”的，只有循环可以被中断。不过，我想要在一个for循环里，让if语句在第一次找到“真”的时候就停止评估。

# Import XML Parser
import xml.etree.ElementTree as ET

# Parse XML directly from the file path
tree = ET.parse('xml file')

# Create iterable item list
items = tree.findall('item')

# Create class for historic variables
class DataPoint:
    def __init__(self, low, high, freq):
        self.low = low
        self.high = high
        self.freq = freq

# Create Master Dictionary and variable list for historic variables
masterDictionary = {}

# Loop to assign variables as dictionary keys and associate their values with them
for item in items:
    thisKey = item.find('variable').text
    thisList = []
    masterDictionary[thisKey] = thisList

for item in items:
    thisKey = item.find('variable').text
    newDataPoint = DataPoint(float(item.find('low').text), float(item.find('high').text), float(item.find('freq').text))
    masterDictionary[thisKey].append(newDataPoint)

diceDictionary = {}
import random
for thisKey in masterDictionary.keys():
    randomValue = random.random()
    diceList = []
    thisList = []
    diceList = masterDictionary[thisKey]
    diceDictionary[thisKey] = thisList
    for i in range(len(diceList)):
        if randomValue <= sum(i.freq for i in diceList[0:i+1]):         
            print 'O', i, 'randomValue', randomValue, 'prob container', sum(i.freq for i in diceList[0:i+1])
            #diceRoll = random.uniform(diceList[i].low, diceList[i].high)
            #diceDictionary[thisKey].append(diceRoll)
        else:
            print 'X', i, 'randomValue', randomValue, 'prob container', sum(i.freq for i in diceList[0:i+1])

在masterDictionary里有两个键，每个键分别包含27个和29个数据点的列表。因此，循环会让i从0到26和从0到28遍历每个键。这很好，但问题在于，当if语句被评估时，一旦找到“真”，接下来的所有项都会一直是“真”。这里是打印输出：

for i in range(len(diceList)):

任何地方出现'X'表示if语句是假的，一旦出现'O'，后面的语句就会一直为真，因为prob容器的大小在不断增加（最大到1.0）。

我想要的是一种方法，让我的if语句在循环里找到第一个“真”后就停止，然后写入字典，再继续外层循环。

任何帮助都非常感谢！

更新：

diceDictionary = {}
x=0 
while x < 3:
    import random
    for thisKey in masterDictionary.keys():
        randomValue = random.random()
        diceList = []
        thisList = []
        diceList = masterDictionary[thisKey]
        diceDictionary[thisKey] = thisList
        for i in range(len(diceList)):
            if randomValue <= sum(i.freq for i in diceList[0:i+1]):         
                print 'O', thisKey, i, 'randomValue', randomValue, 'prob container', sum(i.freq for i in diceList[0:i+1])
                diceRoll = random.uniform(diceList[i].low, diceList[i].high)
                diceDictionary[thisKey].append(diceRoll)
                break
            else:
                print 'X', thisKey, i, 'randomValue', randomValue, 'prob container', sum(i.freq for i in diceList[0:i+1])
    x = x + 1
print diceDictionary

产生：

X inflation 0 randomValue 0.500605733928 prob container 0.0294117647059
X inflation 1 randomValue 0.500605733928 prob container 0.0294117647059
X inflation 2 randomValue 0.500605733928 prob container 0.0294117647059
X inflation 3 randomValue 0.500605733928 prob container 0.0294117647059
X inflation 4 randomValue 0.500605733928 prob container 0.147058823529
X inflation 5 randomValue 0.500605733928 prob container 0.235294117647
X inflation 6 randomValue 0.500605733928 prob container 0.441176470588
O inflation 7 randomValue 0.500605733928 prob container 0.588235294118
X stock 0 randomValue 0.392225720409 prob container 0.0294117647059
X stock 1 randomValue 0.392225720409 prob container 0.0294117647059
X stock 2 randomValue 0.392225720409 prob container 0.0294117647059
X stock 3 randomValue 0.392225720409 prob container 0.0294117647059
X stock 4 randomValue 0.392225720409 prob container 0.0294117647059
X stock 5 randomValue 0.392225720409 prob container 0.0294117647059
X stock 6 randomValue 0.392225720409 prob container 0.0882352941176
X stock 7 randomValue 0.392225720409 prob container 0.0882352941176
X stock 8 randomValue 0.392225720409 prob container 0.0882352941176
X stock 9 randomValue 0.392225720409 prob container 0.117647058824
X stock 10 randomValue 0.392225720409 prob container 0.147058823529
X stock 11 randomValue 0.392225720409 prob container 0.205882352941
X stock 12 randomValue 0.392225720409 prob container 0.264705882353
X stock 13 randomValue 0.392225720409 prob container 0.294117647059
X stock 14 randomValue 0.392225720409 prob container 0.382352941176
O stock 15 randomValue 0.392225720409 prob container 0.441176470588
X inflation 0 randomValue 0.146182475695 prob container 0.0294117647059
X inflation 1 randomValue 0.146182475695 prob container 0.0294117647059
X inflation 2 randomValue 0.146182475695 prob container 0.0294117647059
X inflation 3 randomValue 0.146182475695 prob container 0.0294117647059
O inflation 4 randomValue 0.146182475695 prob container 0.147058823529
X stock 0 randomValue 0.745100497977 prob container 0.0294117647059
X stock 1 randomValue 0.745100497977 prob container 0.0294117647059
X stock 2 randomValue 0.745100497977 prob container 0.0294117647059
X stock 3 randomValue 0.745100497977 prob container 0.0294117647059
X stock 4 randomValue 0.745100497977 prob container 0.0294117647059
X stock 5 randomValue 0.745100497977 prob container 0.0294117647059
X stock 6 randomValue 0.745100497977 prob container 0.0882352941176
X stock 7 randomValue 0.745100497977 prob container 0.0882352941176
X stock 8 randomValue 0.745100497977 prob container 0.0882352941176
X stock 9 randomValue 0.745100497977 prob container 0.117647058824
X stock 10 randomValue 0.745100497977 prob container 0.147058823529
X stock 11 randomValue 0.745100497977 prob container 0.205882352941
X stock 12 randomValue 0.745100497977 prob container 0.264705882353
X stock 13 randomValue 0.745100497977 prob container 0.294117647059
X stock 14 randomValue 0.745100497977 prob container 0.382352941176
X stock 15 randomValue 0.745100497977 prob container 0.441176470588
X stock 16 randomValue 0.745100497977 prob container 0.470588235294
X stock 17 randomValue 0.745100497977 prob container 0.470588235294
X stock 18 randomValue 0.745100497977 prob container 0.529411764706
X stock 19 randomValue 0.745100497977 prob container 0.588235294118
X stock 20 randomValue 0.745100497977 prob container 0.647058823529
O stock 21 randomValue 0.745100497977 prob container 0.764705882353
X inflation 0 randomValue 0.332170052306 prob container 0.0294117647059
X inflation 1 randomValue 0.332170052306 prob container 0.0294117647059
X inflation 2 randomValue 0.332170052306 prob container 0.0294117647059
X inflation 3 randomValue 0.332170052306 prob container 0.0294117647059
X inflation 4 randomValue 0.332170052306 prob container 0.147058823529
X inflation 5 randomValue 0.332170052306 prob container 0.235294117647
O inflation 6 randomValue 0.332170052306 prob container 0.441176470588
X stock 0 randomValue 0.145551106438 prob container 0.0294117647059
X stock 1 randomValue 0.145551106438 prob container 0.0294117647059
X stock 2 randomValue 0.145551106438 prob container 0.0294117647059
X stock 3 randomValue 0.145551106438 prob container 0.0294117647059
X stock 4 randomValue 0.145551106438 prob container 0.0294117647059
X stock 5 randomValue 0.145551106438 prob container 0.0294117647059
X stock 6 randomValue 0.145551106438 prob container 0.0882352941176
X stock 7 randomValue 0.145551106438 prob container 0.0882352941176
X stock 8 randomValue 0.145551106438 prob container 0.0882352941176
X stock 9 randomValue 0.145551106438 prob container 0.117647058824
O stock 10 randomValue 0.145551106438 prob container 0.147058823529
{'inflation': [0.028073642645577577], 'stock': [-0.07388514885974767]}
数据结构 字典操作 循环控制 编程技巧 条件语句 迭代过程 逻辑评估 早期退出

5 个回答

1

那么，什么情况下你能确定这个语句是 True 呢？我原以为可以是“如果最后一个语句是 True”，但在你的示例输出中，最后又回到了 False？

无论如何，考虑在你的 if 语句中添加一个这样的初始条件：

if (you don't already know it's True) and (the condition you currently evaluate):
    <Do calculations>

如果第一部分被判断为 False（也就是说，你已经知道它是 True），那么 Python 就不应该再判断 and 后面的部分了（因为它现在不可能是 True），然后继续执行。你只需要再加一个 else 条件（所以可以把 else 改成 elif），然后在那儿处理它。

注意：这种方法可能有点不太正规，具体取决于你需要做什么来判断你是否已经知道这个语句是 True :\

回答于 2025-04-16 由 Python大师
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2

一种方法是在内部代码中抛出一个异常，然后在for循环里面捕获这个异常，继续执行循环。

回答于 2025-04-16 由 Python大师
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12
    if randomValue <= sum(i.freq for i in diceList[0:i+1]):         
        print 'O', i, 'randomValue', randomValue, 'prob container', sum(i.freq for i in diceList[0:i+1])
        break

Break 这个关键词会结束“最近的一个循环，如果这个循环有可选的 else 部分，它会被跳过。”外面的循环会继续进行下一次循环。所以你并不是在“中断 if 语句”，而是在中断包含这个 if 的循环。在使用 break 之前，你可以把 diceList[0:i+1] 的所有值设置为 diceList[0:len(diceList)+1] 的真值。

回答于 2025-04-16 由 Python大师
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