Question

我有一个Pandas DataFrame，其中有3列，如下所示：

df：

day         min_from  value
2012-10-12    0        0.34
2012-10-12    30       0.36
2012-10-12    60       0.56

df.dtypes

min_from               int64
day                    datetime64[ns]
value                  float64

我正在尝试创建以下内容

date                   value
2012-10-12 00:00:00    0.34
2012-10-12 00:30:00    0.36
2012-10-12 00:60:00    0.56

我尝试过：

def create_dt(row):
    return str(df['day'])+ ' '+ str(timedelta(minutes=df['min_from']))

df['date_time'] = df.apply(create_dt, axis=1)

但是在应用create_dt时出现以下错误：

TypeError: ('unsupported type for timedelta minutes component: Series', 'occurred at index 0')

Answer 1

如果您的day列采用日期时间格式，则可以像这样添加Timedelta，避免调用apply（这可能很慢）：

# In case day is not datetime yet:
df['date'] = pd.to_datetime(df['day']) + pd.to_timedelta(df['min_from'],unit='m')

# Otherwise
df['date'] = df['day'] + pd.to_timedelta(df['min_from'],unit='m')

>>> df
         day  min_from  value                 date
0 2012-10-12         0   0.34  2012-10-12 00:00:00
1 2012-10-12        30   0.36  2012-10-12 00:30:00
2 2012-10-12        60   0.56  2012-10-12 01:00:00

Answer 2

from datetime import timedelta

df['date'] = df.apply(lambda row: row.day + timedelta(minute=row.min_from), axis=1)

将在几分钟内在您的date中创建一个DataFrame列

熊猫从日期和整数创建datetime列

2 个答案: