大熊猫图案匹配数之和

Question

你好，我有一个df，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

我想数一数：

1. Nb群与":“：”=1 (G3)
2. Nb群":“= 2(G1和G4 )
3. Nb群":”=1 (G2)

有人知道吗？我想我应该起诉一个re.sub，然后在熊猫中计算每个Groups的总和？

Answer 1

您可以使用这个来计算使用pd.Series.str.contains，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

Answer 2

使用Series.str.contains作为掩码，然后通过numpy.setdiff1d将DataFrame.loc过滤的值与~或掩码的反向掩码进行比较：

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']

用于计数的Anf获取列表长度：

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