计算熊猫中模式匹配的总和

Question

你好，我有一个df，例如：

Groups COL1
G1   Seq:1
G1   Seq:2
G1   Seq_1
G1   Seq:4
G2   Seq_2
G2   Seq_3
G2   Seq_4
G3   Seq:5
G3   Seq:6
G4   Seq:7
G4   Seq_5

我想数一下：

1. 只有“：” = 1（G3）的Nb组
2. 不仅具有“：” = 2（G1和G4）的Nb组
3. 没有任何“：” = 1（G2）的Nb组

有人不知道吗？我想我应该起诉re.sub并在熊猫中对每个Groups求和？

Answer 1

使用Series.str.contains作为掩码，然后将numpy.setdiff1d的DataFrame.loc过滤值与~或掩码的倒置掩码进行比较：

m = df['COL1'].str.contains(':')

a = np.setdiff1d(df['Groups'], df.loc[~m, 'Groups']).tolist()
print (a)
['G3']

c = np.setdiff1d(df['Groups'], df.loc[m, 'Groups']).tolist()
print (c)
['G2']

b = np.setdiff1d(df.loc[~m, 'Groups'], c).tolist()
print (b)
['G1', 'G4']

计数的Anf获取列表的长度：

print (len(a))
print (len(b))
print (len(c))

Answer 2

您可以使用pd.Series.str.contains进行计数，然后使用GroupBy.all和GroupBy.any

om = df['COL1'].str.contains(':')

one = om.groupby(df['Groups']).all().sum() # 1
two = om.groupby(df['Groups']).any().sum() - one # 2 
# minus one because `any` counts all Trues too so we need 
# subtract groups with all Trues.
three = (~om).groupby(df['Groups']).all().sum() # 1