Question

我想将for循环的输出（0,1,1,2,3）保存到文件中，但我的代码只写入循环的最后一个值（3）。我该如何解决？

#!/usr/bin/python

def fib(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
    return a
for c in range(0, 5):
   print(fib(c))
   file=open("fib.txt","w")
   s = str(fib(c))
   file.write(s+"\n")
#   file.write("%s\n" % fib(c))
   file.close()

Answer 1

尝试这个。

def fib(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
    return a
file=open("fib.txt", "a")
for c in range(0, 5):
   print(fib(c))
   s = str(fib(c))
   file.write(s + "\n")
file.close()

Answer 2

您可能希望了解generators和context managers：

def fib(n):
    a, b = 0, 1
    for i in range(n):
        a, b = b, a + b
        yield a

with open("fib.txt","w") as f:
    for x in fib(5):
        f.write(str(x) + '\n')

Answer 3

好吧，它不仅容易，而且容易得多......：P

使用相同的代码只需在打开时更改文件模式即...

file=open("fib.txt","w") #opens your file in write mode

所以......将其改为

file=open("fib.txt","a") #opens your file in append mode

将以追加模式打开您的文件。

Answer 4

尝试yield代替return

#!/usr/bin/python

def fib(n):
    a, b = 0, 1
    for i in range(0, n):
        a, b = b, a + b
        yield a
for c in range(0, 5):
   print(fib(c))
   file=open("fib.txt","w")
   for s in str(fib(c)):
       file.write(s+"\n")
   file.close()

将循环的输出保存到文件

4 个答案: