python - Pandas 和应用函数来匹配一个字符串

Question

我有一个包含各种链接的 df 列，其中一些包含字符串 "search"。

我想创建一个函数 - 应用于列 - 返回包含 "search" 或 "other" 的列。

我写了一个函数:

search = 'search'
def page_type(x):
if x.str.contains(search):
    return 'Search'
else:
    return 'Other'   

df['link'].apply(page_type)

但它给了我这样的错误:

AttributeError: 'unicode' object has no attribute 'str'

我想我在调用 str.contains() 时遗漏了一些东西。

Answer 1

我想你需要numpy.where :

df = pd.DataFrame({'link':['search','homepage d','login dd', 'profile t', 'ff']})

print (df)
         link
0      search
1  homepage d
2    login dd
3   profile t
4          ff

search = 'search'
profile = 'profile'
homepage = 'homepage'
login = "login"

def page_type(x):
    if search in x:
        return 'Search'
    elif profile in x:
        return 'Profile'
    elif homepage in x:
        return 'Homepage'
    elif login in x:
        return 'Login'
    else:
        return 'Other'  

df['link_new'] = df['link'].apply(page_type)

df['link_type'] = np.where(df.link.str.contains(search),'Search', 
                  np.where(df.link.str.contains(profile),'Profile', 
                  np.where(df.link.str.contains(homepage), 'Homepage', 
                  np.where(df.link.str.contains(login),'Login','Other')))) 


print (df)
         link  link_new link_type
0      search    Search    Search
1  homepage d  Homepage  Homepage
2    login dd     Login     Login
3   profile t   Profile   Profile
4          ff     Other     Other

时间:

#[5000 rows x 1 columns]
df = pd.DataFrame({'link':['search','homepage d','login dd', 'profile t', 'ff']})
df = pd.concat([df]*1000).reset_index(drop=True)

In [346]: %timeit df['link'].apply(page_type)
1000 loops, best of 3: 1.72 ms per loop

In [347]: %timeit np.where(df.link.str.contains(search),'Search', np.where(df.link.str.contains(profile),'Profile', np.where(df.link.str.contains(homepage), 'Homepage', np.where(df.link.str.contains(login),'Login','Other'))))
100 loops, best of 3: 11.7 ms per loop