【发布时间】:2017-02-20 19:57:04
【问题描述】:
我正在从字段中捕获数据并将其存储为
data.append(newdata)
然后将每条完整记录存储为:
list_of_rows.append(data)
然后我尝试保存到 csv
data_mod = [[item] for item in list_of_rows]
with open("./hotels.csv", "wb") as outfile:
writer = csv.writer(outfile)
for row in data_mod:
writer.writerow(row)
outfile.close()
但是当我将它加载到 csv 中时,所有内容都会保存到第一个字段中。如何正确分解?
编辑
每一行看起来像
[[u'Staybridge Suites London - Vauxhall', '\nTushar K\n', '\nIlford\n', 0, u'2 reviews', '5 of 5 stars', '29 September 2016', u'\nHome comes at staybridge........it nice with stay bridge.....awesome ambiance, kitchen, rooms, break fast area.............\nEverything is at place.....\nTalking about people of stay bridge... they all are very much cooperative, kind, best service people I have ever saw, meet...... its absolutely fantastic with stay bridge..........love u guysss.....\n']]
【问题讨论】:
-
为什么要以二进制模式打开文件?无关,但您可以使用
writer.writerows(data_mod) -
我不知道我正在以二进制打开 - 该行应该是什么?
writer.writerows(data_mod)仍然不允许我在单独的字段中加载到 csv;一切都进入第一列 -
"wb"b-> 二进制模式,你只需要w。data_mod长什么样子? -
添加了一个示例输出行
-
列表中的列表?
标签: python-2.7 csv beautifulsoup