python-在Pandas DataFrame中构建复杂的子集

我正在使用GroupBy,但仍然需要一些帮助.假设我有一个带有列Group的DataFrame,为对象提供了组编号,一些参数R和球坐标RA和Dec.这是一个模拟DataFrame：

df = pd.DataFrame({ 
    'R'    : (-21.0,-21.5,-22.1,-23.7,-23.8,-20.4,-21.8,-19.3,-22.5,-24.7,-19.9),
    'RA': (154.362789,154.409301,154.419191,154.474165,154.424842,162.568516,8.355454,8.346812,8.728223,8.759622,8.799796),
    'Dec': (-0.495605,-0.453085,-0.481657,-0.614827,-0.584243,8.214719,8.355454,8.346812,8.728223,8.759622,8.799796),
    'Group': (1,1,1,1,1,2,2,2,2,2,2) 
})

我想建立一个包含每个组的“最亮”对象的选择,即具有最小R(或最大绝对值,因为Ris为负值)和该组中3个最接近的对象(所以我在每个对象中保留4个对象)组-我们可以假设不存在小于4个对象的组).

我们在这里假设我们定义了以下功能：

#deg to rad
def d2r(x):
    return x * np.pi / 180.0

#rad to deg
def r2d(x):
    return x * 180.0 / np.pi

#Computes separation on a sphere
def calc_sep(phi1,theta1,phi2,theta2):
    return np.arccos(np.sin(theta1)*np.sin(theta2) + 
                     np.cos(theta1)*np.cos(theta2)*np.cos(phi2 - phi1) )

并且两个对象之间的间隔由r2d(calc_sep(RA1,Dec1,RA2,Dec2))给出,其中RA1作为第一个对象的RA,依此类推.

我不知道如何使用GroupBy实现这一目标…

解决方法:

您在这里可以做的是构建一个更具体的帮助器函数,该函数将应用于每个“子框架”(每个组).

实际上,GroupBy只是一种设施,它创建了类似(组ID,DataFrame)对的迭代器,并且当您调用.groupby().apply时,将对每个函数应用一个函数. (有很多详细信息,如果您有兴趣,请参阅here,以获取有关内部的一些详细信息.)

因此,在定义了三个基于NumPy的函数之后,还需要定义：

def sep_df(df, keep=3):
    min_r = df.loc[df.R.argmin()]
    RA1, Dec1 = min_r.RA, min_r.Dec
    sep = r2d(calc_sep(RA1,Dec1,df['RA'], df['Dec']))
    idx = sep.nsmallest(keep+1).index
    return df.loc[idx]

然后只需应用,您将获得一个MultiIndex DataFrame,其中第一个索引级别是该组.

print(df.groupby('Group').apply(sep_df))
              Dec  Group     R         RA
Group                                    
1     3  -0.61483      1 -23.7  154.47416
      2  -0.48166      1 -22.1  154.41919
      0  -0.49561      1 -21.0  154.36279
      4  -0.58424      1 -23.8  154.42484
2     8   8.72822      2 -22.5    8.72822
      10  8.79980      2 -19.9    8.79980
      6   8.35545      2 -21.8    8.35545
      9   8.75962      2 -24.7    8.75962

穿插着一些评论：

def sep_df(df, keep=3):
    # Applied to each sub-Dataframe (this is what GroupBy does under the hood)

    # Get RA and Dec values at minimum R
    min_r = df.loc[df.R.argmin()]  # Series - row at which R is minimum
    RA1, Dec1 = min_r.RA, min_r.Dec  # Relevant 2 scalars within this row

    # Calculate separation for each pair including minimum R row
    # The result is a series of separations, same length as `df`
    sep = r2d(calc_sep(RA1,Dec1,df['RA'], df['Dec']))

    # Get index values of `keep` (default 3) smallest results
    # Retain `keep+1` values because one will be the minimum R
    # row where separation=0
    idx = sep.nsmallest(keep+1).index

    # Restrict the result to those 3 index labels + your minimum R
    return df.loc[idx]

对于速度,如果结果仍然适合您,则将consider passing sort=False更改为GroupBy.

标签：pandas-groupby,pandas,dataframe,split-apply-combine,python
来源： https://codeday.me/bug/20191025/1930495.html