我無法在斐波那契數列 python 中加強術語

Question

我想打印 fibonacci seq 的每個第四項，直到 n 項，但是我很困惑為什么這段代碼不起作用，我首先有一個函數來計算 n 項，然后是另一個函數來迭代和每第四項步進。

# Program to display the Fibonacci sequence up to n-th term
def fibo_sequence(n):

# first two terms
    n1, n2 = 0, 1
    count = 0

# check if the number of terms is valid
    if n <= 0:
        print("Please enter a positive integer")
# if there is only one term, return n1
    elif n == 1:
        print("Fibonacci sequence upto",n,":")
        print(n1)
# generate fibonacci sequence
    else:
        print("Fibonacci sequence:")
        while count < n:
            print(n1)
            nth = n1 + n2
            # update values
            n1 = n2
            n2 = nth
            count += 1


def fibo_skipper(n):
    #fibo_sequence(n)
    for term in fibo_sequence(n):
        for i in range(0, n, 4):
            print(i)





n = 8  #Enter the nth term here

fibo_skipper(n) # Print the print every fourth item of the Fibonacci sequence upto n

Answer 1

問題是您在最低級別使用print ，因此無法從更高級別過濾任何內容。 如果您希望能夠過濾，則低fibo_sequence(n)函數不應打印任何內容，而應返回一個序列。 一個簡單的方法是使用列表：

# Program to display the Fibonacci sequence up to n-th term
def fibo_sequence(n):
    seq = []

# first two terms
    n1, n2 = 0, 1
    count = 0

# check if the number of terms is valid
    if n <= 0:
        print("Please enter a positive integer")
# if there is only one term, return n1
    elif n == 1:
        print("Fibonacci sequence upto",n,":")
        seq.append(n1)
# generate fibonacci sequence
    else:
        print("Fibonacci sequence:")
        while count < n:
            seq.append(n1)
            nth = n1 + n2
            # update values
            n1 = n2
            n2 = nth
            count += 1
    return seq

但它仍然不會遵循良好的封裝實踐，因為相同的函數顯示消息並構建序列。 您應該只構建序列（可選的空列表）並讓調用者決定在返回列表的長度為 0 或 1 時顯示什么。

如果你想更進一步，你應該考慮構建一個生成器，一次yield一個值，因此避免存儲它們。

Answer 2

注意：我從假設您的fibo_sequence實際上返回一個sequence開始寫了一個答案......我的錯。

顯然，很自然的情況是您有某種斐波那契計算（即返回一個序列，使用生成器）返回一個可處理的輸出，正如許多解決方案所建議的那樣。

無論如何，你仍然可以做你喜歡做的事情，即使有打印到屏幕的功能，臨時重定向sys.stdout

from io import StringIO
import sys

def fibo_skipper(n, step = 4):
    old_stdout = sys.stdout
    sys.stdout = StringIO()
    fibo_sequence(n)
    lines = sys.stdout.getvalue().split("\n")[1:] # [1:] is to skip first line
    sys.stdout = old_stdout
    # slice operator will split your lines every n elements
    print(lines[::step])


fibo_skipper(100)

重要的是要強調，我不是在建議這種方法（這超出了您原來的問題），但是，在您無法完全控制被調用方法（在您的情況下為fibo_sequence ）的情況下，這仍然可以讓您到達需要。

作為旁注，我認為您可以更緊湊的else語句：

else:
    print("Fibonacci sequence:")
    while count < n:
        print(n1)
        # update values
        n1, n2 = n2, n1 + n2
        count += 1

Answer 3

在您當前的代碼結構中，您應該在打印之前檢查第n-th元素是否可以被 4 整除，並且您可以使用模數運算符%檢查

...
# generate fibonacci sequence
    else:
        print("Fibonacci sequence:")
        while count < n:
            if count % 4 == 0:
                print(n1)
            nth = n1 + n2
            # update values
            n1 = n2
            n2 = nth
            count += 1

通過這種方式，您仍然可以計算系列並僅打印每個第 4 個元素。

如果您希望跳過大小作為參數，您可以將其傳遞給您的函數：

def fibo_sequence(n, skip):
...
# generate fibonacci sequence
    else:
        print("Fibonacci sequence:")
        while count < n:
            if count % skip == 0:
                print(n1)
            nth = n1 + n2
            # update values
            n1 = n2
            n2 = nth
            count += 1

請注意，此代碼結構對更改的魯棒性較差。

Answer 4

您將在 main 中調用 skipper 函數。 skipper 函數調用序列函數，該函數返回斐波那契數列。 for 循環以 4 的間隔打印這些數字。

# Program to display the Fibonacci sequence up to n-th term
def fibo_sequence(n):   
    l = []
    # first two terms
    n1, n2 = 0, 1
    count = 0

    # check if the number of terms is valid
    if n <= 0:
        print("Please enter a positive integer")
    # if there is only one term, return n1
    elif n == 1:
        print("Fibonacci sequence upto",n,":")
        l.append(n1)
    # generate fibonacci sequence
    else:
        print("Fibonacci sequence:")
        while count < n:
            l.append(n1)
            nth = n1 + n2
            # update values
            n1 = n2
            n2 = nth
            count += 1
    return l

def fibo_skipper(n):
    l = fibo_sequence(n)
    for i in range(0, n, 4):
        print(l[i])


n = int(input())
fibo_skipper(n)

Answer 5

我的第一個答案是垃圾，給你：

def fibo_sequence(n):

    n1, n2 = 0, 1
    count = 0
    seq = []

   if n <= 0:
       print("Please enter a positive integer")
   
   elif n == 1:
       print("Fibonacci sequence upto",n,":")
       print(n1)
   else:
       print("Fibonacci sequence:")
       while count < n:
           print(n1)
           seq.append(n1)
           nth = n1 + n2
           # update values
           n1, n2 = n2, nth
           count += 1

   if len(seq) != 0:
       return seq

def fibo_skipper(n):
    print("Every 4th Number in Fibo up to n:  ")
    seq = fibo_sequence(n)
    for el in seq:
        print(seq[el*4])

n = 10  
fibo_skipper(n)