我有一个多索引数据框dfu：
open high low close
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.030
09:48:00 25.90 25.90 25.760 25.760
09:51:00 26.00 26.00 25.985 25.985
2016-11-29 09:30:00 24.98 24.98 24.98 24.9800
09:33:00 25.00 25.00 24.99 24.9900
09:35:00 25.33 25.46 25.33 25.4147
我想创建一个新列，[＆＃39; closeScaled＆＃39;]，它是通过使用[＆＃39;的当前级别= 0值的第一行来执行函数foo计算的。打开＆＃39;]列和当前行[＆＃39;关闭＆＃39;]作为参数。我怀疑解决方案会涉及到：
dfu['closeScaled']=dfu.apply(lambda x: foo(*get first row of current date*[0],x[3]))
我似乎无法弄清楚获取当前级别的第一行= 0 部分。
如果foo是：
def foo(firstOpen,currentClose):
return (currentClose / firstOpen)
然后我希望closeScaled列包含（截断为4位小数）：
open high low close closeScaled
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.030 1.0000
09:48:00 25.90 25.90 25.760 25.760 0.9896
09:51:00 26.00 26.00 25.985 25.985 0.9982
2016-11-29 09:30:00 24.98 24.98 24.98 24.9800 1.0000
09:33:00 25.00 25.00 24.99 24.9900 1.0004
09:35:00 25.33 25.46 25.33 25.4147 1.0174
答案 0 :(得分：2)
您可以将div创建的groupby Series与transform first和round分开：
print (dfu.groupby(level=0)['open'].transform('first'))
Date Time
2016-11-28 09:43:00 26.03
09:48:00 26.03
09:51:00 26.03
2016-11-29 09:30:00 24.98
09:33:00 24.98
09:35:00 24.98
Name: open, dtype: float64
dfu['closeScaled'] = dfu.close.div(dfu.groupby(level=0)['open'].transform('first')).round(4)
print (dfu)
open high low close closeScaled
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.0300 1.0000
09:48:00 25.90 25.90 25.760 25.7600 0.9896
09:51:00 26.00 26.00 25.985 25.9850 0.9983
2016-11-29 09:30:00 24.98 24.98 24.980 24.9800 1.0000
09:33:00 25.00 25.00 24.990 24.9900 1.0004
09:35:00 25.33 25.46 25.330 25.4147 1.0174
如果需要将浮点值截断为4位小数：
首先按10000复制，转换为int并除以10000。
dfu['closeScaled'] = dfu.close.div(dfu.groupby(level=0)['open'].transform('first'))
.mul(10000).astype(int).div(10000)
print (dfu)
open high low close closeScaled
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.0300 1.0000
09:48:00 25.90 25.90 25.760 25.7600 0.9896
09:51:00 26.00 26.00 25.985 25.9850 0.9982
2016-11-29 09:30:00 24.98 24.98 24.980 24.9800 1.0000
09:33:00 25.00 25.00 24.990 24.9900 1.0004
09:35:00 25.33 25.46 25.330 25.4147 1.0174
#http://stackoverflow.com/a/783927/2901002
def truncate(f, n):
'''Truncates/pads a float f to n decimal places without rounding'''
s = '{}'.format(f)
if 'e' in s or 'E' in s:
return '{0:.{1}f}'.format(f, n)
i, p, d = s.partition('.')
return '.'.join([i, (d+'0'*n)[:n]])
dfu['closeScaled'] = dfu.close.div(dfu.groupby(level=0)['open'].transform('first'))
.apply(lambda x: truncate(x,4)).astype(float)
print (dfu)
open high low close closeScaled
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.0300 1.0000
09:48:00 25.90 25.90 25.760 25.7600 0.9896
09:51:00 26.00 26.00 25.985 25.9850 0.9982
2016-11-29 09:30:00 24.98 24.98 24.980 24.9800 1.0000
09:33:00 25.00 25.00 24.990 24.9900 1.0004
09:35:00 25.33 25.46 25.330 25.4147 1.0174
答案 1 :(得分：2)
使用groupby + apply + lambda
df.groupby(level=0).apply(
lambda df: df.assign(closeScaled=df.close.div(df.open.iloc[0]).round(4))
)
open high low close closeScaled
Date Time
2016-11-28 09:43:00 26.03 26.03 26.030 26.0300 1.0000
09:48:00 25.90 25.90 25.760 25.7600 0.9896
09:51:00 26.00 26.00 25.985 25.9850 0.9983
2016-11-29 09:30:00 24.98 24.98 24.980 24.9800 1.0000
09:33:00 25.00 25.00 24.990 24.9900 1.0004
09:35:00 25.33 25.46 25.330 25.4147 1.0174