我有df：
CU Parameters 1 2 3
379-H Output Energy, (Wh/h) 0.045 0.055 0.042
349-J Output Energy, (Wh/h) 0.001 0.003 0
625-H Output Energy, (Wh/h) 2.695 1.224 1.272
626-F Output Energy, (Wh/h) 1.381 1.494 1.3
我想创建两个单独的dfs，通过在0级（CU）上对索引进行分组来获取列值的平均值：
df1：（379-H和625-H）
Parameters 1 2 3
Output Energy, (Wh/h) 1.37 0.63 0.657
df2 :(其余的）
Parameters 1 2 3
Output Energy, (Wh/h) 0.69 0.74 0.65
我可以通过分组级别1获得所有使用的均值：
df = df.apply(pd.to_numeric, errors='coerce').dropna(how='all').groupby(level=1).mean()
但如何根据0级对这些进行分组？
解决方案：
lightsonly = ["379-H", "625-H"]
df = df.apply(pd.to_numeric, errors='coerce').dropna(how='all')
mask = df.index.get_level_values(0).isin(lightsonly)
df1 = df[mask].groupby(level=1).mean()
df2 = df[~mask].groupby(level=1).mean()
答案 0 :(得分：2)
考虑数据框df，其中假设CU和Parameters位于索引中。
1 2 3
CU Parameters
379-H Output Energy, (Wh/h) 0.045 0.055 0.042
349-J Output Energy, (Wh/h) 0.001 0.003 0.000
625-H Output Energy, (Wh/h) 2.695 1.224 1.272
626-F Output Energy, (Wh/h) 1.381 1.494 1.300
然后我们可以根据第一级值是否在列表['379-H', '625-H']中的真值进行分组。
m = {True: 'Main', False: 'Rest'}
l = ['379-H', '625-H']
g = df.index.get_level_values('CU').isin(l)
df.groupby(g).mean().rename(index=m)
1 2 3
Rest 0.691 0.7485 0.650
Main 1.370 0.6395 0.657
答案 1 :(得分：2)
使用get_level_values + isin获取True和False索引，然后rename获取dict d = {True: '379-H and 625-H', False: 'the rest'}
df.index = df.index.get_level_values(0).isin(['379-H', '625-H'])
df = df.mean(level=0).rename(d)
print (df)
1 2 3
the rest 0.691 0.7485 0.650
379-H and 625-H 1.370 0.6395 0.657
：< / p>
dfs对于单独的mask= df.index.get_level_values(0).isin(['379-H', '625-H'])
df1 = df[mask].mean().rename('379-H and 625-H').to_frame().T
print (df1)
1 2 3
379-H and 625-H 1.37 0.6395 0.657
df2 = df[~mask].mean().rename('the rest').to_frame().T
print (df2)
1 2 3
the rest 0.691 0.7485 0.65
，也可以使用mean：
numpy使用DataFrame构造函数的另一个a1 = df[mask].values.mean(axis=0)
#alternatively
#a1 = df.values[mask].mean(axis=0)
df1 = pd.DataFrame(a1.reshape(-1, len(a1)), index=['379-H and 625-H'], columns=df.columns)
print (df1)
1 2 3
379-H and 625-H 1.37 0.6395 0.657
解决方案：
select Outlets.OutletName,
avg(datediff(ms, Orders.OrderDate, ReceivedOrders.ReceivingDate)) / 60000 as Receive
from dbo.Orders
inner join dbo.Outlets on dbo.Orders.OutletID = dbo.Outlets.OutletID
inner join dbo.ReceivedOrders on dbo.Orders.OrderID = dbo.ReceivedOrders.OrderID
group by dbo.Outlets.OutletName
答案 2 :(得分：1)
*** 1 failure is detected in the test module "cmPTP Test Suite"
# cmPTP-manager-test -l all --run_test=comms_rproc_interface/constructor
Running 1 test case...
Entering test module "cmPTP Test Suite"
interface_test.cpp(17): Entering test suite "comms_rproc_interface"
interface_test.cpp(19): Entering test case "constructor"
interface_test.cpp(23): info: check master_pty != -1 has passed
interface_test.cpp(24): info: check grantpt(master_pty) != -1 has passed
interface_test.cpp(25): info: check unlockpt(master_pty) != -1 has passed
interface_test.cpp(32): info: check slave_pty != -1 has passed
interface_test.cpp(72): info: check in_data.size() == bytes_written has passed
interface_test.cpp(77): error: in "comms_rproc_interface/constructor": check bytes_read == bytes_written has failed [0 != 4]
interface_test.cpp(19): Leaving test case "constructor"; testing time: 15282us
interface_test.cpp(17): Leaving test suite "comms_rproc_interface"; testing time: 15931us
Leaving test module "cmPTP Test Suite"; testing time: 16879us
*** 1 failure is detected in the test module "cmPTP Test Suite"