Generating .xml files based on .csv files

Question

This reads in data from a .csv file and generates an .xml file with respect to the .csv data. Can you spot any parts where re-factoring can make this code more efficient?

import csv from xml.etree.ElementTree import Element, SubElement, Comment, tostring from xml.etree.ElementTree import ElementTree import xml.etree.ElementTree as etree def main(): reader = read_csv() generate_xml(reader) def read_csv(): with open ('1250_12.csv', 'r') as data: return list(csv.reader(data)) def generate_xml(reader): root = Element('Solution') root.set('version','1.0') tree = ElementTree(root) head = SubElement(root, 'DrillHoles') head.set('total_holes', '238') description = SubElement(head,'description') current_group = None i = 0 for row in reader: if i > 0: x1,y1,z1,x2,y2,z2,cost = row if current_group is None or i != current_group.text: current_group = SubElement(description, 'hole',{'hole_id':"%s"%i}) collar = SubElement (current_group, 'collar',{'':', '.join((x1,y1,z1))}), toe = SubElement (current_group, 'toe',{'':', '.join((x2,y2,z2))}) cost = SubElement(current_group, 'cost',{'':cost}) i+=1 def indent(elem, level=0): i = "\n" + level*" " if len(elem): if not elem.text or not elem.text.strip(): elem.text = i + " " if not elem.tail or not elem.tail.strip(): elem.tail = i for elem in elem: indent(elem, level+1) if not elem.tail or not elem.tail.strip(): elem.tail = i else: if level and (not elem.tail or not elem.tail.strip()): elem.tail = i indent(root) tree.write(open('holes1.xml','w')) 

Answer 1

Minor stuff:

• Remove the last import - etree is not used anywhere.
• Merge the two first imports

Possibly speed-improving stuff:

• Avoid converting the csv.reader output before returning unless absolutely necessary.
• Skip indent unless the output must be readable by a human with a non-formatting editor.
• If you need to indent the output, existing solutions are probably very efficient.
• Use reader.next() to skip the header line in generate_xml, then you don't need to keep checking the value of i.

Answer 2

Don't use something like for elem in elem at some point with larger for loops you will miss that elem is different variable before and in/after the for loop:

for subelem in elem: indent(subelem, level+1) if not subelem.tail or not elem.tail.strip(): subelem.tail = i 

Since indent(subelem...) already sets the tail, you probably do not need to do that again.