Question

我有一个字符串列（＆＃39; b＆＃39;），并希望获得与同一列中的子字符串类似的字符串。例如，在下面的数据框列中＆＃39; b＆＃39;世界是helloworld的子串，ness是伟大的子串。我想在列表中获取字符串世界和ness。能否请你提出解决方案。

     a           b
0  test       world
1  teat  helloworld
2   gor         bye
3   jhr   greatness
4   fre        ness

列表中的所需输出

listofsubstrings
Out[353]: ['world', 'ness']

Answer 1

您可以使用：

from itertools import product

#get unique values only
b = df.b.unique()
#create all combination
df1 = pd.DataFrame(list(product(b, b)), columns=['a', 'b'])
#filtering
df1 = df1[df1.apply(lambda x: x.a in x.b, axis=1) & (df1.a != df1.b)]
print (df1)
        a           b
1   world  helloworld
23   ness   greatness

print (df1.a.tolist())
['world', 'ness']

交叉加入的替代解决方案：

b = df.b.unique()
df['tmp'] = 1
df1 = pd.merge(df[['b','tmp']],df[['b','tmp']], on='tmp')
df1 = df1[df1.apply(lambda x: x.b_x in x.b_y, axis=1) & (df1.b_x != df1.b_y)]
print (df1)
      b_x  tmp         b_y
1   world    1  helloworld
23   ness    1   greatness

print (df1.b_x.tolist())
['world', 'ness']

Answer 2

这可能对您有用：

df_cross = pd.DataFrame(data=np.asarray(df.b) + " " + df.b[:,None], columns=df.b)
df_indicator = df_cross.applymap(lambda x: x.split()[0] in x.split()[1])
df_indicator.sum(axis=0)[lambda x: x>1].index

Out[231]: Index([u'world', u'ness'], dtype='object')

Answer 3

我们可以创建一个真值数组，而行索引是列标题的子字符串。

l = df.b.dropna().values  # grab values from b
# double comprehension
a = np.array([[j in i for i in l] for j in l])
# of course strings are sub-strings of themselves
# lets ignore them by making the diagonal `False`
np.fill_diagonal(a, False)

# find the indices where the array is `True`
i, j = np.where(a)

l[i].tolist()

['world', 'ness']

更好的imo

s = pd.Series(l[i], l[j])
s

helloworld    world
greatness      ness
dtype: object

Python Pandas - 在同一列中查找元素（substring）

3 个答案: