python - 如何对我的 DataFrame 中的值进行分组和计数？

Question

我有这个数据框:

df = pd.DataFrame(columns=["App","Feature1", "Feature2","Feature3",
                           "Feature4","Feature5",
                           "Feature6","Feature7","Feature8"], 
                  data=[["SHA",0,0,1,1,1,0,1,0],
                        ["LHA",1,0,1,1,0,1,1,0],
                        ["DRA",0,0,0,0,0,0,1,0],
                        ["FRA",1,0,1,1,1,0,1,1],
                        ["BRU",0,0,1,0,1,0,0,0],
                        ["PAR",0,1,1,1,1,0,1,0],
                        ["AER",0,0,1,1,0,1,1,0],
                        ["SHE",0,0,0,1,0,0,1,0]])

更新:(抱歉，我错误地表述了预期结果)

我想计算每个特征值 1 出现的次数:

Features   Count
Feature1   6
Feature2   7
...

我试过这个:

df.groupBy("App").count()

但我没有得到预期的输出。

Answer 1

使用:

#remove column App, compare and get sum of Trues
a0 = df.drop('App', 1).eq(0).sum()
#a0 = df.set_index('App').eq(0).sum()

#alternative with select only Feature columns
#a0 = df.filter(like='Feature').eq(0).sum()

#alternative with select all columns without first
a0 = df.iloc[:, 1:].eq(0).sum()

print (a0)
Feature1    6
Feature2    7
Feature3    2
Feature4    2
Feature5    4
Feature6    6
Feature7    1
Feature8    7
dtype: int64

与 1 比较类似:

a1 = df.drop('App', 1).eq(1).sum()
#a1 = df.set_index('App').eq(1).sum()

#alternative
#a1 = df.filter(like='Feature').eq(1).sum()
#alternative
a1 = df.iloc[:, 1:].eq(1).sum()

print (a1)
Feature1    2
Feature2    1
Feature3    6
Feature4    6
Feature5    4
Feature6    2
Feature7    7
Feature8    1
dtype: int64

全部与value_counts一起:

a = df.drop('App', 1).apply(pd.value_counts).T.add_prefix('count_')
print (a)
          count_0  count_1
Feature1        6        2
Feature2        7        1
Feature3        2        6
Feature4        2        6
Feature5        4        4
Feature6        6        2
Feature7        1        7
Feature8        7        1

或者列表理解:

L = [df[x].value_counts() for x in df.columns.difference(['App'])]
a = pd.concat(L, 1).T.add_prefix('count_')
print (a)
          count_0  count_1
Feature1        6        2
Feature2        7        1
Feature3        2        6
Feature4        2        6
Feature5        4        4
Feature6        6        2
Feature7        1        7
Feature8        7        1