我要转换以下XML文件:
<data>
<level_1 name="employment">
<level_2 name="sub-employment">
<indicator>ind1</indicator>
<indicator>ind2</indicator>
</level_2>
<level_2 name="sub-employment2">
<indicator>ind3</indicator>
</level_2>
</level_1>
<level_1 name="health">
<level_2 name="sub-health">
<level_3 name="sub-sub-health">
<indicator>ind4</indicator>
</level_3>
</level_2>
</level_1>
</data>
进入熊猫数据框,结果类似于:
level_1 level_2 level_3 indicator
0 employment sub-employment None ind1
1 employment sub-employment None ind2
2 employment sub-employment2 None ind3
3 health sub-health sub-sub-health ind4
在将xml.etree.cElementTree导入为et并将熊猫导入为pd之后,我使用了以下代码:
def getvalueofnode(node):
""" return node text or None """
return node.text if node is not None else None
def main():
""" main """
parsed_xml = et.parse("tree.xml")
dfcols = ['level_1', 'level_2', 'level_3', 'indicator']
df_xml = pd.DataFrame(columns=dfcols)
for node in parsed_xml.getroot():
name = node.attrib.get('name')
level_2 = node.find('level_2')
level_3 = node.find('level_3')
indicator = node.find('indicator')
df_xml = df_xml.append(
pd.Series([name, getvalueofnode(level_2), getvalueofnode(level_3),
getvalueofnode(indicator)], index=dfcols),
ignore_index=True)
print(df_xml)
main()
但是我得到了错误的结果:
level_1 level_2 level_3 indicator
0 employment \n None None
1 health \n None None
我在做什么错了?
答案 0 :(得分:1)
定义以下功能,创建祖先字典, 从 node 向上:
def parNames(node, root):
names = {}
while True:
node = parentMap[node]
if node is root:
return names
names[node.tag] = node.attrib['name']
稍后将需要它。它使用 parentMap 字典,该字典将 即将创建。
读取您的输入文件:
tree = et.parse('tree.xml')
root = tree.getroot()
实际处理应从创建父级地图开始 -字典,每个节点返回其父节点:
parentMap = {}
for parent in root.iter():
for child in parent:
parentMap[child] = parent
要为您的DataFrame创建源数据,请运行:
rows = []
for it in root.iter('indicator'):
row = parNames(it, root)
row[it.tag] = it.text
rows.append(row)
此循环创建词典列表(每行数据)。 每行(一个字典)包含:
下一步是创建DataFrame:
df2 = pd.DataFrame(rows).fillna('').sort_index(axis=1)
唯一要做的步骤是将 indicator 列移到最后一个位置:
df2 = df2.reindex(df2.columns.drop('indicator')
.append(pd.Index(['indicator'])),axis=1)