Saving zip list to csv in Python

落爺英雄遲暮 提交于 2019-12-05 15:40:07

writer.writerows expects a sequence of values for writing a single row into the CSV file.

Using your original code:

import csv
res =[{'date': '2015/01/01 00:00', 'v': 96.5}, {'date': '2015/01/01 00:01', 'v': 97.0}, {'date': '2015/01/01 00:02', 'v': 93.75}, {'date': '2015/01/01 00:03', 'v': 96.0}, {'date': '2015/01/01 00:04', 'v': 94.5}]
csvfile = "example.csv"
with open(csvfile, "w") as output:
  writer = csv.writer(output, lineterminator='\n')
  for line in res:
    date = line['date']
    value = line['v']
    writer.writerow([date, value])

Since I find csv.DictWriter not transparent in what its doing, I would recommend doing the following:

with open(csvfile, "w") as output:
    output.write(';'.join(list(res[0].keys()))+"\n")
    [output.write(';'.join(list(map(str, r.values())))+"\n") for r in res]

You can switch to using Python's DictWriter for this. You can pass a list of column headers, this ensures that the order of the columns in the output is what you require. Only columns in this list are written to the your output file:

import csv

zip_list = [
    {'date': '2015/01/01 00:00', 'v': 96.5},
    {'date': '2015/01/01 00:01', 'v': 97.0},
    {'date': '2015/01/01 00:02', 'v': 93.75},
    {'date': '2015/01/01 00:03', 'v': 96.0},
    {'date': '2015/01/01 00:04', 'v': 94.5}]

csvfile = "/home/stm/PycharmProjects/isbak_trafik/example.csv"

with open(csvfile, "wb") as output:
    writer = csv.DictWriter(output, fieldnames=['date', 'v'])
    writer.writeheader()
    writer.writerows(zip_list)

This would produce the following output:

date,v
2015/01/01 00:00,96.5
2015/01/01 00:01,97.0
2015/01/01 00:02,93.75
2015/01/01 00:03,96.0
2015/01/01 00:04,94.5
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