如何逐行将itertools.izip写入txt文件？

Question

我的代码

import itertools
import os

with open("base.txt") as fv, open("new2.txt", 'r') as fi, open('sortedf.txt','w') as fs:
    vel = (line.strip() for line in fv)
    ind = (int(line.strip()) for line in fi)
    z = itertools.izip(ind, vel) # sort according to ind
    # itertools.izip(vel, ind) # sort according to vel
    for i, v in sorted(z):
        fs.write(str(v))

我把所有东西放在一行

2.900000e+032.900000e+032.900000e+032.900000e+032.

当我换到  fs.write('\n'.join(str(v)))

然后我得到了

2
.
9
0
0
0
0
0
e
+
0
32
.
9
0
0
0
0
0
e
+
0
32
.

如何逐行获取正确的值？

Answer 1

只需改变

for i, v in sorted(z):
    fs.write(str(v))

到

for i,v in sorted(z):
    print(v, file=fs)

由于\n
的end="\n"默认参数，

• print会自动添加

• 适用于任何数据类型。无需str(v)

Answer 2

请尝试以下

fs.writelines(map(lambda x: x[1], sorted(z)))

为什么以下声明失败

fs.write('\n'.join(str(v)))

这里将v转换为列表，并在其上应用连接。为清晰起见，请查看以下示例

>>> sam = 'hello'
>>> '-'.join(sam)
'h-e-l-l-o'

那么如何使用fs.write？

1. 写作：fs.write（v）
2. 现在添加换行符：fs.write（'\ n'）

很少有建议：

import os # load important modules first

from itertools import izip # makes processing faster, if you only need izip 


with open("base.txt") as fv, open("new2.txt", 'r') as fi, open('sortedf.txt','w') as fs:
    vel = [line.strip() for line in fv] # use square braces
    ind = int(line.strip()) for line in fi] # use square braces
    z = izip(ind, vel) # sort according to ind
    for i, v in sorted(z):
        fs.write(v)
        fs.write('\n') # adding line break