在循環之外保持Python條件？

因此，我在Python中瞭解到了這個雙區搜索算法，我正在使用它來查找的近似數字的平方根。至於算法，它工作正常，它的侷限性，但是，這是我在的JavaScript經常做的，是封裝內的變量條件，所以它更容易閱讀，如：在循環之外保持Python條件？

var isGreaterThanFive = num > 5; 
if(isGreaterThanFive && otherConditions...)

雖然在JavaScript的作品這樣做當嘗試在Python中執行此操作時，絕對沒問題，我的程序似乎進入無限循環。這是我在JavaScript代碼：

function sqrtOf (x) { 
    var min = 0 
    var max = x 
    var epsylon = 0.001 
    var guess = (max + min)/2 
    var guessNumber = 0; 

    //I created these two so it is easier to understand 
    var notCloseEnough = Math.abs(Math.pow(guess, 2) - x) >= epsylon; 
    var stillPlausible = guess <= x; 

    while (notCloseEnough && stillPlausible) { 
    guessNumber += 1 

    if(Math.abs(Math.pow(guess, 2)) > x) { 
     max = guess; 
    } else { 
     min = guess; 
    } 

    guess = (max + min)/2; 
    } 

    return guess; 
} 

console.log(sqrtOf(25)); // 5

在Python

現在：

def sqrtOf (x) 
    minVal = 0 
    maxVal = x 
    epsylon = 0.001 
    guess = (maxVal + minVal)/2.0 
    guessNumber = 0; 

    notCloseEnough = abs(guess ** 2 - x) >= epsylon 
    stillPlausible = guess <= x 

    while notCloseEnough and stillPlausible: 
    guessNumber += 1 

    if abs(guess ** 2) > x: 
     maxVal = guess 
    else: 
     minVal = guess 

    guess = (maxVal + minVal)/2.0 

    return guess 

print sqrtOf(25)

來源

2016-12-15 zavjs

當你有和賦值時，'='右邊的表達式在賦值給變量之前被求值。所以'stillPlausibe'在循環開始的時候等於'True'或'False'，並且你永遠不會做任何事情來改變它。 –

你的javascript代碼有同樣的問題... – Izkata

當你寫

notCloseEnough = abs(guess ** 2 - x) >= epsylon

你評估的聲明abs(guess ** 2 - x) >= epsylon並指派其結果notCloseEnough。這種計算不會再次發生，因爲您之後在代碼中碰巧引用了它的結果。

如果你想重新評估事物，你需要一個函數對象。你可以定義一個內部函數，它可以看到外部函數的局部變量。

def sqrtOf (x): 
    minVal = 0 
    maxVal = x 
    epsylon = 0.001 
    guess = (maxVal + minVal)/2.0 
    guessNumber = 0 

    def notCloseEnough(): 
    return abs(guess ** 2 - x) >= epsylon 

    def stillPlausible(): 
    return guess <= x 

    while notCloseEnough() and stillPlausible(): 
    guessNumber += 1 

    if abs(guess ** 2) > x: 
     maxVal = guess 
    else: 
     minVal = guess 

    guess = (maxVal + minVal)/2.0 

    return guess 

print sqrtOf(25)

來源

2016-12-15 16:09:11 tdelaney

這也不錯，stillPlausible讓我感到困惑。輸入應該在循環之前得到驗證。

def sqrtOf(x): 
    minVal = 0 
    maxVal = x 
    epsylon = 0.001 
    guess = (maxVal + minVal)/2.0 
    guessNumber = 0; 
    def trytrytry(): 
    while True: 
     yield (abs(guess ** 2 - x) < epsylon, guess - x < epsylon) 

    for (closeEnough, stillPlausible) in trytrytry(): 
    if closeEnough or not stillPlausible: 
     break 
    guessNumber += 1 

    if abs(guess ** 2) > x: 
     maxVal = guess 
    else: 
     minVal = guess 

    guess = (maxVal + minVal)/2.0 

    return guess 

print sqrtOf(25)

來源

2016-12-15 16:19:51

在循環之外保持Python條件？

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