所以我有一些看起来像这样的代码:
measure_list=[]
for file in file_list:
with open(file,"r") as read_data:
y=read_data.read()
if "Part: ABCD" in y:
for line in m:
if "Measure X:" in line:
measure_list.append(line)
elif "Measure Y:" in line:
measure_list.append(line)
final=''.join(measure_list).replace("Measure","\nMeasure")
print(final)
(这最后一部分只是帮助组织输出)
因此,这部分代码将打开一组文件,并扫描每个文件以查看其是否为“ Part:ABCD”,如果是,它将拉出“ Measure X:”和“ Measure Y:”行并添加他们到measure_list。我的问题是我有很多文件,并且会有多个“ Part:ABCD”文件。在使用.join()加入列表之后,输出将如下所示:
Part: ABCD
Measure X: (Numbers)
Measure Y: (Numbers)
Measure X: (Numbers)
Measure Y: (Numbers)
Measure X: (Numbers)
Measure Y: (Numbers)
我的问题是,是否有人知道组织输出的方式,使其看起来像这样:
Part: ABCD
Measure X: (Numbers)
Measure X: (Numbers)
Measure X: (Numbers)
Measure Y: (Numbers)
Measure Y: (Numbers)
Measure Y: (Numbers)
任何帮助表示赞赏。
您measure_list
看起来像这样:
['Measure X: 1000', 'Measure Y: 2000', 'Measure X: 100' , 'Measure Y: 900']
如果是的话,您可以简单地将其排序measure_list
:
measure_list.sort()
然后您将拥有:
['Measure X: 100', 'Measure X: 1000', 'Measure Y: 2000', 'Measure Y: 900']
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我来说两句