转换为“日期时间”类型的问题：“小时必须在 0..23”

Question

这些是我的 csv 文件中的一些示例行：

10/10/1949 20:30,san marcos,tx,us,cylinder,2700,45 minutes,"This event took place in early fall around 1949-50. It occurred after a Boy Scout meeting in the Baptist Church. The Baptist Church sit",4/27/2004,29.8830556,-97.9411111
10/10/1949 21:00,lackland afb,tx,,light,7200,1-2 hrs,"1949 Lackland AFB&#44 TX.  Lights racing across the sky & making 90 degree turns on a dime.",12/16/2005,29.38421,-98.581082
10/10/1955 17:00,chester (uk/england),,gb,circle,20,20 seconds,"Green/Orange circular disc over Chester&#44 England",1/21/2008,53.2,-2.916667
10/10/1956 21:00,edna,tx,us,circle,20,1/2 hour,"My older brother and twin sister were leaving the only Edna theater at about 9 PM&#44...we had our bikes and I took a different route home",1/17/2004,28.9783333,-96.6458333

完整的 csv 文件在这里。

我将其加载到数据框中。在列名中'datetime'，我有格式'object'。我试图将类型转换'object'为这样的类型'datetime'：

df['datetime'] = pd.to_datetime(df.datetime)

结果我收到了这个错误：

ValueError: hour must be in 0..23

任何帮助将不胜感激！

Answer 1

很显然的问题是24:00，解决办法是Series.str.split，date人民共同转换to_datetime，并time通过to_timedelta与加在一起：

print (df)
           datetime
0  10/10/1949 20:30
1  10/10/1949 21:00
2  10/10/1955 17:00
3  10/10/1956 24:00

df[['date','time']] = df['datetime'].str.split(expand=True)
df['datetime'] = (pd.to_datetime(df.pop('date'), format='%d/%m/%Y') + 
                  pd.to_timedelta(df.pop('time') + ':00'))
print (df)
             datetime
0 1949-10-10 20:30:00
1 1949-10-10 21:00:00
2 1955-10-10 17:00:00
3 1956-10-11 00:00:00