Python 文本格式处理问题

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This topic created in 2546 days ago, the information mentioned may be changed or developed.

文本如下

Storage 1:
total storage = 204699 MB
free storage = 195091 MB
Storage 2:
total storage = 204699 MB
free storage = 200661 MB

如何实现最终结果为

Storage_1_total_storage = 204699 MB
Storage_1_free_storage = 195091 MB
Storage_2_total_storage = 204699 MB
Storage_2_free_storage = 200661 MB

11 replies • 2019-05-24 22:14:14 +08:00

zhila

May 24, 2019

我的处理比较繁琐，坐等楼下大佬回答。

csdreamdong

May 24, 2019

笨方法：根据换行 split 后。遍历。再根据特定的条件，去拼接。。、

V2tizen

May 24, 2019

第一反应是正则不知道楼下大佬有没有简单方法

fisher335

May 24, 2019 via Android

cinfigparser 模块吧这个明显符合 windows 配资文件的格式然后直接解析成字典

ysc3839

May 24, 2019

如果 Storage total storage 这些都是固定的话，用正则可以这么写：
`Storage (\d+):\ntotal storage = (.+)\nfree storage = (.+)`

如果不固定的话还是别用正则了，逐行遍历可能没那么麻烦。

dairoot

May 24, 2019

for line in file:
if line[-1:] == ':':
prefix = line[:-1].replace(' ', '_')
else:
print (prefix+'_'+line.replace(' ', '_', 1))

Cooky

May 24, 2019 via Android

for line in text :
if line.startswith('Storage') :
s = line.strip().replace( ' ', '_' )［:-1 ］
else :
k,v = line.split('=')
k = k.strip().replace( ' ', '_' )
line = '%s_%s = %s' % ( s, k, v )
out.write(line)
差不多就这样

Cooky

May 24, 2019 via Android

@Cooky sad，空格被吃掉了

zeraba

May 24, 2019 via Android

configparser +1

omph

May 24, 2019

```python3
for line in open("new.txt", "r"):
line = line.rstrip()
if line.startswith("Storage "):
storage = line.rstrip(':')
else:
key, val = map(lambda x:x.strip(), line.split('='))
new_key = f"{storage}_{key}".replace(' ', '_')
print(f"{new_key} = {val}")
```

jianghu52

May 24, 2019

6 楼的思路不错啊。学到了。replace 竟然有第三个参数。以前一直不知道