簡單的Python隨機數猜測器

我剛剛寫了一個猜測隨機數的快速腳本。雖然劇本的最後一行'恭喜你贏了'等不在運行。同樣在一些解釋器中，它給出了一個錯誤'運行腳本失敗'。簡單的Python隨機數猜測器

import random 
attempts = 0 
secret_number = random.randint(1,100) 
isCorrect = False 
guess = int(input("Take a guess: ")) 

while secret_number != guess: 
    if guess < secret_number: 
     print("Higher...") 
     guess = int(input("Take a guess: ")) 
     attempts+= 1 
    elif guess > secret_number: 
     print("Lower...") 
     guess = int(input("Take a guess: ")) 
     attempts+= 1 
    else: 
     print("\nYou guessed it! The number was " ,secret_number)

來源

2017-09-08 Ibo Cetinkaya

如果初始'guess'是正確的，循環條件是立即虛假，不執行循環。 – DyZ

else行永遠不會執行，因爲while循環條件在正確的猜測中得到驗證並退出循環。你可以刪除else和un-indent那一行，這樣它在顯示後顯示 – iffy

將最後一行移出while循環 – chowsai

你應該在while循環中打印出來，因爲當secret_number == guess時你的循環將會中斷。

while ...: 
    if: 
     ... 
    elif: 
     ... 

print "You guessed it! The number was {0}".format(secret_number)

來源

2017-09-08 07:32:19 DLavid

我對腳本做了一些修改。需要注意的是其功能是通過執行流程，沒有什麼特別的控制語句（如continue，break）或明確的干預調控是必要的，使得它在邏輯上是一致的：

import random 


secret_number = random.randint(1, 100) 
guess = None 
attempts = 0 

while secret_number != guess: 
    guess = int(input("Take a guess: ")) 
    attempts += 1 

    if secret_number == guess: 
     print("\nYou guessed it! The number was ", secret_number) 
    elif guess < secret_number: 
     print("Higher...") 
    elif guess > secret_number: 
     print("Lower...")

我改變：

的isCorrect變量。我刪除了它，因爲它沒有被使用;
變量初始化的方式guess。 while語句前爲None;
while循環內的條件語句;
在while循環內從用戶讀取輸入的順序。

我建議直觀瞭解腳本的工作原理。

我試圖運行它，並收到了以下的輸出：

[email protected]:~$ python3.6 ./test.py 
Take a guess: 70 
Higher... 
Take a guess: 80 
Higher... 
Take a guess: 90 
Higher... 
Take a guess: 95 
Lower... 
Take a guess: 94 
Lower... 
Take a guess: 93 
Lower... 
Take a guess: 92 
Lower... 
Take a guess: 91 

You guessed it! The number was 91

來源

2017-09-08 07:38:03 none

簡單的Python隨機數猜測器

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