我使用输出到具有以下格式的文本文件的空间数据:
COMPANY NAME
P.O. BOX 999999
ZIP CODE , CITY
+99 999 9999
23 April 2013 09:27:55
PROJECT: Link Ref
--------------------------------------------------------------------------------
--------------------------------------------------------------------------------
Design DTM is 30MB 2.5X2.5
Stripping applied to design is 0.000
Point Number Easting Northing R.L. Design R.L. Difference Tol Name
3224808 422092.700 6096059.380 2.520 -19.066 -21.586 --
3224809 422092.200 6096059.030 2.510 -19.065 -21.575 --
<Remainder of lines>
3273093 422698.920 6096372.550 1.240 -20.057 -21.297 --
Average height difference is -21.390
RMS is 21.596
0.00 % above tolerance
98.37 % below tolerance
End of Report
如图所示,文件有页眉和页脚。数据由空格分隔,但列之间的数量不相等。
我需要的是以逗号分隔的文件,其中包含东距、北距和差值。
我想避免手动修改数百个大文件,我正在编写一个小脚本来处理这些文件。这是我目前所拥有的:
#! /usr/bin/env python
import csv,glob,os
from itertools import islice
list_of_files = glob.glob('C:/test/*.txt')
for filename in list_of_files:
(short_filename, extension )= os.path.splitext(filename)
print short_filename
file_out_name = short_filename + '_ed' + extension
with open (filename, 'rb') as source:
reader = csv.reader( source)
for row in islice(reader, 10, None):
file_out= open (file_out_name, 'wb')
writer= csv.writer(file_out)
writer.writerows(reader)
print 'Created file: '+ file_out_name
file_out.close()
print 'All done!'
问题:
如何让以“Point number”开头的行成为输出文件中的标题?我正在尝试用 DictReader 代替读写器位,但无法正常工作。
用分隔符“,”写入输出文件确实有效,但在每个空格处写了一个逗号,让我的输出文件中有太多空列。我该如何避免这种情况?
如何删除页脚?
最佳答案
我发现你的代码有问题,你正在为每一行创建一个新的 writer
;所以你最终只会得到最后一个。
您的代码可能是这样的,不需要 CSV 读取器或写入器,因为它足够简单,可以被解析为简单文本(如果您有文本列、转义字符等,就会出现问题)。
def process_file(source, dest):
found_header = False
for line in source:
line = line.strip()
if not header_found:
#ignore everything until we find this text
header_found = line.starswith('Point Number')
elif not line:
return #we are done when we find an empty line, I guess
else:
#write the needed columns
columns = line.split()
dest.writeline(','.join(columns[i] for i in (1, 2, 5)))
for filename in list_of_files:
short_filename, extension = os.path.splitext(filename)
file_out_name = short_filename + '_ed' + extension
with open(filename, 'r') as source:
with open(file_out_name. 'w') as dest:
process_file(source, dest)
关于python - Python中的CSV文件处理,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16275498/