类型错误：strptime() 参数 1 必须是 str，而不是 Period答案

【问题标题】：TypeError: strptime() argument 1 must be str, not Period类型错误：strptime() 参数 1 必须是 str，而不是 Period
【发布时间】：2019-12-03 21:08:35
【问题描述】：

我有这个数据框。

import pandas as pd
from datetime import datetime
df = pd.DataFrame({'id': [11,22,33,44,55], 
                   'name': ['A','B','C','D','E'], 
                   'timestamp': [1407617838,965150022,1158531592,1500701864,965149631]})

df
   id name timestamp
0  11    A      2014
1  22    B      2000
2  33    C      2006
3  44    D      2017
4  55    E      2000

df['timestamp'] = pd.to_datetime(df['timestamp'], unit='s')
df['timestamp'] = df['timestamp'].dt.to_period('Y')
y1 = df['timestamp'].iloc[0]
y2 = df['timestamp'].iloc[1]
d1 = datetime.strptime(y1, "%Y")
d2 = datetime.strptime(y2, "%Y")
diff = abs((d2 - d1).days)
print(diff)

我已将时间戳转换为实际日期并获取年份。我想要两个取前两行时间戳之间的差异。 例如 (abs (2014-2000) = 4)

【问题讨论】：

标签： python pandas

【解决方案1】：

如果您通过dt acessor of timeseries 计算年份，您会得到整数（而不是“Period”对象）：

df['timestamp'] = pd.to_datetime(df['timestamp'], unit='s')
df['timestamp'] = df['timestamp'].dt.year
y1 = df['timestamp'].iloc[0]
y2 = df['timestamp'].iloc[1]
# d1 = datetime.strptime(y1, "%Y") <- No need to recast to datetime!
# d2 = datetime.strptime(y2, "%Y")
diff = abs((y2 - y1))
print(diff)
>>> 14

如您所见，我评论这两行是您试图将年份重铸为 datetime 对象。这是有原因的吗？根据您的问题，我假设您想要年数的差异。如果您想要时间戳之间的确切天数，那么应该这样做：（无需转换和重新转换）：

df['timestamp'] = pd.to_datetime(df['timestamp'], unit='s')
y1 = df['timestamp'].iloc[0]
y2 = df['timestamp'].iloc[1]
diff = abs((y2 - y1).days)
print(diff)
>>> 5122

【讨论】：