Question

我正在尝试将输出列表打印到文件中作为下面的输出，但它一直给我错误。

示例输入

样本输入的position.dat内容

[1, 0]
[0, 2]
[-4, -4]
[-64, 0]
[65536, 0]
[4294967296, 0]
[1329227995784915872903807060280344576, 1329227995784915872903807060280344576]

这是我的代码：

infile = open("position.dat", "w")
def B(n):
    direction=[[1,0],[0,1],[-1,0],[0,-1]] #right, up, left, down
    start=[0,0]
    x=start[0]
    y=start[1]    
    if n>50000:
        return "Do not enter input that is larger than 50000"
    elif n==0:   #base case
        return [1, 0]
    elif n==1:
        return [1, 1]

    elif n%2==0 and n%4!=0:  #digit start from n=2 and every 4th n digit
        x=0            # start from second digit (n) x=0 for every 4th digit
        y=((-4)**(n//4))*2 


    elif n%4==0:      #print out every 4 digits n
        y=0  #every 4digit of y will be 0 start from n=0
        x=(-4)**(n//4)  #the pattern for every 4th digits

    elif n>3 and n%2 !=0 and n%4==1: #start from n=1 and for every 4th digit
        x=(-4)**(n//4)
        y=(-4)**(n//4)

    elif n%4==3 and n%2 != 0:  #start from n=3
        y=((-4)**(n//4))*2
        x=((-4)**(n//4))*-2
    return [int(x),int(y)]     #return result


print(B(0))    # print the input  onto python shell
print(B(2))
print(B(5))
print(B(12)) 
print(B(32))
print(B(64))
print(B(241))
print(B(1251))
#Please also input the integers below for printing it on the the file

infile.write(B(0)+'\n')   # these keep giving me error
infile.write(B(2)+'\n')   



infile.close()

是否可以使用括号将列表打印到文件上？

Answer 1

__str__的{{1}}方法可以为您完成此操作。即使用

list

Answer 2

你可以使用类似的东西：

with open ("temp.txt","wt") as fh:
   fh.write("%s\n" % str([0,1]))

使用格式化原语
明确地将您的元素转换为字符串

或者，使用个人格式：

fh.write("[%s]\n" %(','.join(map (str,B(0))))

Answer 3

您无法使用write直接撰写列表。但是，您可以生成原始字符串：

s0 = "[" + ", ".join([str(x) for x in B(0)]) + "]"
infile.write(s0 + "\n")

如何使用括号将列表打印到python中的文件中

3 个答案: