我无法提高斐波那契数列的条件 python

Question

我想打印斐波那契数列的每第四项直到 n 项但是我很困惑为什么这段代码不起作用，我首先有一个函数来计算 n 项然后另一个函数迭代和步进每个第四学期.

# 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

我的第一个答案是垃圾，给你：

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)

Answer 2

注意：我写了一个答案，假设你的fibo_sequence实际上返回了一个sequence......我的错。

显然，自然情况是您有某种斐波那契计算（即返回序列，使用生成器）返回可处理的输出，正如许多解决方案所建议的那样。

无论如何，你仍然可以做你喜欢的事情，即使有打印到屏幕的功能，暂时重定向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)

重要的是要强调，我并不是在建议这种方法（这超出了您最初的问题），而是在您无法完全控制被调用方法的情况下（在您的情况下，fibo_sequence ), 这仍然可以让你到达你需要的地方。

作为旁注，我认为您可以将 else 语句更紧凑：

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

Answer 3

在您当前的代码结构中，您应该在打印之前检查 n-th 元素是否可以被 4 整除，并且您可以使用取模运算符 %

进行检查

...
# 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

这样您仍然可以计算系列并仅打印每个第 4 个元素。

如果你想将跳过大小作为参数，你可以将它传递给你的函数：

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

请注意，此代码结构对更改的鲁棒性较差。

Answer 4

你会在main中调用船长函数。船长函数调用序列函数，其中 returns 斐波那契数列。 for 循环以 4 为间隔打印这些数字。

# 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

问题是您在最低级别使用 print，因此无法从更高级别过滤任何内容。如果你想过滤，low fibo_sequence(n) 函数不应该打印任何东西，而应该 return 一个序列。一个简单的方法是使用列表：

# 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

但它仍然不会遵循良好的封装实践，因为相同的函数显示消息并构建序列。如果 returned 列表的长度为 0 或 1，您应该只构建序列（可选的空列表）并让调用者决定显示什么。

如果您想更进一步，您应该考虑构建一个生成器，该生成器一次 yield 一个值，因此可以避免存储它们。

我无法提高斐波那契数列的条件 python

I'm not able to step up terms in fibonacci sequence python

python

fibonacci