Python：“爆发”； for循环中的if语句 [英] Python: "breaking out" of if statement inside a for loop

问题描述

我知道一个人不能中断 if语句，只能从循环中进行，但是，我试图从概念上阻止if语句在for循环中第一次找到 true后才对其求值。

I understand that one cannot "break" an if statement and only from a loop, however, I'm trying to conceptually stop an if statement from evaluating after it finds a "true" the first time when it's inside a for loop.

# 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])

有是masterDictionary中的两个键，每个键分别包含27个和29个数据点的列表。因此，对于范围i中的i的循环

There are two keys in the masterDictionary and each one contains a list of 27 and 29 data points respectively. Therefore, the loop

for i in range(len(diceList)):

每个键的i值分别为0-26和0-28。很好，但是评估if语句时的问题是，一旦找到它，则随后的所有以下范围项都将变为true。打印输出如下：

will run i from 0 - 26 and 0 - 28 for each key. This is great, but the problem when the if statement is evaluated is that once it is found, it will subsequently be true for all of the following range items. Here is the print output:

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

在任何地方都有 X表示if语句为假，并且一旦启动 O，其余语句将始终为true，这是因为大小增加了问题容器（最高1.0）。

Anywhere there is an 'X' means that the if statement was false, and once an 'O' starts, the rest of the statements will always be true because of the increasing size of the prob container (up to 1.0).

我正在寻找的一种方法是，告诉我循环中的if语句一旦找到第一个真实语句就停止，然后写到字典中，然后再继续执行外部循环。

What I am looking for is a way to tell my if statement inside the loop to stop once it finds the first true statement, then write to dictionary, and then continue the outer loop again.

任何帮助表示赞赏！

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]}

推荐答案

    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 将终止最近的封闭循环，如果该循环有一个循环，则跳过可选的else子句。外循环将继续下一次迭代。因此，您不是在打破if，而是将if括在其中。在打破之前，您可以只设置 diceList [0：i + 1] 中的所有值为 diceList [0：len（diceList）+1] 为true。

Break will terminate "the nearest enclosing loop, skipping the optional else clause if the loop has one." The outer loop will just continue with the next iteration. So you are not "breaking the if" but the loop the if is enclosed in. Before the break, you can just set all values from diceList[0:i+1] to diceList[0:len(diceList)+1] to true.

这篇关于Python：“爆发”； for循环中的if语句的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！