我正在尝试计算用户在 Instagram 上的发帖频率。因此，我建立了他们最近发布日期的列表。像这样:
['01-23-2019', '01-19-2019', '01-12-2019', '12-30-2018', '12-28-2018', '12-20-2018', '11-21-2018', '11-09-2018', '10-26-2018', '10-12-2018', '09-30-2018', '09-16-2018', '09-06-2018', '08-31-2018', '08-15-2018', '08-12-2018', '08-09-2018', '07-30-2018', '07-27-2018', '07-24-2018', '07-20-2018', '07-17-2018', '07-14-2018', '07-08-2018', '07-06-2018', '06-30-2018', '06-26-2018', '06-13-2018', '06-08-2018', '06-06-2018', '05-28-2018', '05-21-2018', '05-19-2018', '05-11-2018', '05-08-2018', '05-03-2018', '05-01-2018', '04-12-2018', '04-05-2018', '03-31-2018', '03-27-2018', '03-10-2018', '03-06-2018', '02-25-2018', '02-21-2018', '02-18-2018', '02-16-2018', '02-11-2018', '02-06-2018', '02-03-2018']
理想情况下，我想获得发布日期之间的平均天数。所以我最终得到了一个频率数:例如'用户每 n 天发布一次'。
我从 JSON 代码中获取时间戳并将其转换为如下所示的可读内容:
import datetime
#prepare timestamp to calculate frequency
taken_on = post_details['taken_at_timestamp']
readable_post_date = datetime.datetime.fromtimestamp(taken_on).strftime('%m-%d-%Y')
post_dates.append(readable_post_date)
我应该如何最好地处理这个以获得小数结果？
请您参考如下方法:
只需将日期之间的差异相加并除以总数即可。为了简单起见
from datetime import datetime, timedelta
dates = ['01-23-2019', '01-19-2019', '01-12-2019', '12-30-2018', '12-28-2018', '12-20-2018', '11-21-2018', '11-09-2018', '10-26-2018', '10-12-2018', '09-30-2018', '09-16-2018', '09-06-2018', '08-31-2018', '08-15-2018', '08-12-2018', '08-09-2018', '07-30-2018', '07-27-2018', '07-24-2018', '07-20-2018', '07-17-2018', '07-14-2018', '07-08-2018', '07-06-2018', '06-30-2018', '06-26-2018', '06-13-2018', '06-08-2018', '06-06-2018', '05-28-2018', '05-21-2018', '05-19-2018', '05-11-2018', '05-08-2018', '05-03-2018', '05-01-2018', '04-12-2018', '04-05-2018', '03-31-2018', '03-27-2018', '03-10-2018', '03-06-2018', '02-25-2018', '02-21-2018', '02-18-2018', '02-16-2018', '02-11-2018', '02-06-2018', '02-03-2018']
sorted_dates = sorted(datetime.strptime(d, '%m-%d-%Y') for d in dates)
time_difference = timedelta(0)
counter = 0
for i in range(1, len(sorted_dates), 1):
counter += 1
time_difference += sorted_dates[i] - sorted_dates[i-1]
frequency = time_difference / counter
print(frequency.days) # 7 days
