将 zip 列表保存到 Python 中的 csv

Question

如何将下面的 zip 列表写入 python 中的 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

由于我发现 csv.DictWriter 其功能并不透明，因此我建议您执行以下操作：

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]

Answer 2

writer.writerows 需要一系列值来将单行写入 CSV 文件。

使用您的原始代码：

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])

Answer 3

为此您可以改用 Python 的 DictWriter。您可以传递列 headers 的列表，这可确保输出中列的顺序符合您的要求。仅将此列表中的列写入您的输出文件：

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

将 zip 列表保存到 Python 中的 csv

Saving zip list to csv in Python

python

csv

zip

export

export-to-csv