Python create an xml file for every row in excel-CodePudding

I have the following code which creates a xml file from each row in excel. I want to write the xml files in a new folder (currently generates files in same folder as python), how do i change the "output.write()" statement? I am new in programming, started with python 2 weeks ago.

wb_data

df_data is a list of dictionaries.

[{'Nume': 'Ciuica', 'Prenume': 'Larisa Bianca', 'Data': '30/01/1991 00:00:00', 'Cod': 789454, 'Localitate': 'Targu Jiu'},
 {'Nume': 'Balasa', 'Prenume': 'Valentin', 'Data': '14/09/1989 00:00:00', 'Cod': 215487, 'Localitate': 'Bucuresti'},
 {'Nume': 'Vinatoru', 'Prenume': 'Costin', 'Data': '06/12/1980 00:00:00', 'Cod': 254896, 'Localitate': 'Titu'}]

Code:

import xml.etree.ElementTree as Et
import pandas as pd

wb_data = pd.read_excel("test.xlsx", sheet_name="Data", header=0)
wb_xml = pd.read_excel("test.xlsx", sheet_name="XML", header=0)

df_data = wb_data.to_dict("records")

lista_xml = []
index_lista = 0

for xml in wb_xml["Denumire_xml"]:
    lista_xml.append(xml)

for row in df_data:
    document = Et.Element("Document")
    for key, value in row.items():
        element = Et.SubElement(document, key)
        element.text = str(value)
    output = Et.ElementTree(document)
    Et.indent(output)
    output.write(lista_xml[index_lista])
    index_lista = index_lista   1

CodePudding user response：

Simply use absolute path especially with os.path.join to concatenate folder and file name:

import os
...

xml_file_path = "/path/to/my/xml/outputs"

output.write(
    os.path.join(xml_file_path, lista_xml[index_lista])
)

However, consider the new DataFrame.to_xml() to export data frame rows to XML files:

import os
import pandas as pd

xml_file_path = "/path/to/my/xml/outputs"

wb_data = pd.read_excel("test.xlsx", sheet_name="Data", header=0)
wb_xml = pd.read_excel("test.xlsx", sheet_name="XML", header=0)

# LIST COMPREHENSION
lista_xml = [xml for xml in wb_xml["Denumire_xml"]]

# SERIES TO LIST CONVERSION
lista_xml = wb_xml["Denumire_xml"].tolist()

# GROUP BY INDEX (I.E., ROW) WITH LOOP NUMBER
for n, (i, sub_df) in enumerate(wb_data.groupby(level=0)):
    sub_df.to_xml(
        os.path.join(xml_file_path, lista_xml[n])
    )