python - 在 Python 中将 zip 列表保存到 csv

Question

如何在 python 中将压缩列表下方写入 csv 文件？

[{'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}

我有这个错误:

_csv.Error: sequence expected

我的代码在这里:

import csv
res = zip_list
csvfile = "/home/stm/PycharmProjects/isbak_trafik/example.csv"

with open(csvfile, "w") as output:
    writer = csv.writer(output, lineterminator='\n')
    writer.writerows(res)

Answer 1

为此，您可以切换到使用 Python 的 DictWriter。您可以传递列标题列表，这可确保输出中列的顺序符合您的要求。只有此列表中的列会写入您的输出文件:

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)

这将产生以下输出:

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