在多列上使用pandas groupby函数(Use pandas groupby function on multiple columns)

在多列上使用pandas groupby函数(Use pandas groupby function on multiple columns)

我有一个类似于此的DataFrame：

Key Departure Species1 Species2 Status 1 R Carlan Carlan D 1 R Scival Carex C 2 R Carlan Scival D 2 R Scival Bougra C 3 D Carlan Carlan D 3 D Scival Scival C

我想计算每个独特Species1在给定Departure的出现次数和C的D的Status

我想要的输出是：

Species1 RD RC DD DC Carlan 2 NaN 1 NaN Scival NaN 2 NaN 1

I have a DataFrame similar to this:

Key Departure Species1 Species2 Status 1 R Carlan Carlan D 1 R Scival Carex C 2 R Carlan Scival D 2 R Scival Bougra C 3 D Carlan Carlan D 3 D Scival Scival C

I want to count the occurrences of each unique Species1 for a given Departure and Status of D of C

My desired output is:

Species1 RD RC DD DC Carlan 2 NaN 1 NaN Scival NaN 2 NaN 1

最满意答案

使用pandas.crosstab（）方法。 一行代码：

pd.crosstab(df.Species1, [df.Departure, df.Status])

结果表：

如果你结合@ dermen的'梳子'专栏，

df['comb'] = df.Departure + df.Status pd.crosstab(df.Species1, df.comb)

你会得到：

如果你真的想要那些'NaN'，只需要添加.replace('0', np.nan) ，就像这样（假设import numpy as np已经完成了import numpy as np ）：

pd.crosstab(df.Species1, df.comb).replace('0', np.nan)

Use the pandas.crosstab() method. A single line of code:

pd.crosstab(df.Species1, [df.Departure, df.Status])

The resulting table:

If you combine with @dermen's 'comb' column,

df['comb'] = df.Departure + df.Status pd.crosstab(df.Species1, df.comb)

you'll get:

If you really want those 'NaN', just tack on a .replace('0', np.nan), like so (assuming an import numpy as np has already been done):

pd.crosstab(df.Species1, df.comb).replace('0', np.nan)