Question

我试图比较2个字符串列表的相似性，并将它们显示在pandas数据框中以供检查;所以我使用1个列表作为索引，另一个作为列列表。然后我想计算它们的“Levenshtein相似性”（比较两个单词之间的相似性的函数）。

我试图通过使用apply map来实现这一目标，它将进入每个单元格，并将单元格索引与单元格列进行比较。但我怎么能这样做？或者可能会有一些更简单的方法？

things = ['car', 'bike', 'sidewalk', 'eatery']
action = ['walking', 'caring', 'biking', 'eating']
matrix = pd.DataFrame(index = things, columns = action)

def lev(x):
    x = Levenshtein.distance(x.index, x.column)  
matrix.applymap(lev)

到目前为止，我使用了以下（下面），但我发现它笨拙而且很慢

matrix = pd.DataFrame(data = [action for i in things], index = things, columns = action)
for i, values in matrix.iterrows():
    for j, value in enumerate(values):
        matrix.ix[i,j] = Levenshtein.distance(i, value)

Answer 1

我认为您可以使用apply，而对于列值，请使用.name：

def lev(x):
    #replace your function
    return x.index + x.name
a = matrix.apply(lev)
print (a)
                  walking          caring          biking          eating
car            carwalking       carcaring       carbiking       careating
bike          bikewalking      bikecaring      bikebiking      bikeeating
sidewalk  sidewalkwalking  sidewalkcaring  sidewalkbiking  sidewalkeating
eatery      eaterywalking    eaterycaring    eaterybiking    eateryeating

编辑：

如果需要一些arithemtic操作使用broadcasting：

a = pd.DataFrame(matrix.index.values + matrix.columns.values[:,None], 
                 index=matrix.index, 
                 columns=matrix.columns)
print (a)
             walking       caring           biking         eating
car       carwalking  bikewalking  sidewalkwalking  eaterywalking
bike       carcaring   bikecaring   sidewalkcaring   eaterycaring
sidewalk   carbiking   bikebiking   sidewalkbiking   eaterybiking
eatery     careating   bikeeating   sidewalkeating   eateryeating

或者：

a = pd.DataFrame(matrix.index.values + matrix.columns.values[:, np.newaxis], 
                 index=matrix.index, 
                 columns=matrix.columns)
print (a)
             walking       caring           biking         eating
car       carwalking  bikewalking  sidewalkwalking  eaterywalking
bike       carcaring   bikecaring   sidewalkcaring   eaterycaring
sidewalk   carbiking   bikebiking   sidewalkbiking   eaterybiking
eatery     careating   bikeeating   sidewalkeating   eateryeating

Answer 2

您可以通过“嵌套apply”来做到这一点，如下所示：

things = ['car', 'bike', 'sidewalk', 'eatery']
action = ['walking', 'caring', 'biking', 'eating']
matrix = pd.DataFrame(index=things, columns=action)
matrix.apply(lambda x: pd.DataFrame(x).apply(lambda y: LD(x.name, y.name), axis=1))

输出：

          walking  caring  biking  eating
car             6       3       6       5
bike            6       5       3       5
sidewalk        7       8       7       8
eatery          6       5       6       3

这里的调用pd.DataFrame(x)是因为x是Series对象，而Series.apply与applymap类似，它不携带{{1} }或index信息。

Pandas - 在applymap期间检索每个元素的行名和列名

2 个答案: