Question

我有一个像这样的pandas数据框：

a      b      c
foo    bar    baz
bar    foo    baz
foobar barfoo baz

我在python中定义了以下函数：

def somefunction (row):
    if row['a'] == 'foo' and row['b'] == 'bar':
        return 'yes'
    return 'no'

它完美无缺。但我需要对if函数进行一些小调整，以考虑partial string次匹配。

我尝试了几种组合，但我似乎无法让它发挥作用。我收到以下错误：

("'str' object has no attribute 'str'", 'occurred at index 0')

Iv尝试的功能是：

def somenewfunction (row):
    if row['a'].str.contains('foo')==True and row['b'] == 'bar':
        return 'yes'
    return 'no'

Answer 1

使用contains表示布尔掩码，然后使用numpy.where：

m = df['a'].str.contains('foo') & (df['b'] == 'bar')
print (m)
0     True
1    False
2    False
dtype: bool

df['new'] = np.where(m, 'yes', 'no')
print (df)
        a       b    c  new
0     foo     bar  baz  yes
1     bar     foo  baz   no
2  foobar  barfoo  baz   no

或者，如果需要检查列b以查找子字符串：

m = df['a'].str.contains('foo') & df['b'].str.contains('bar')
df['new'] = np.where(m, 'yes', 'no')
print (df)
        a       b    c  new
0     foo     bar  baz  yes
1     bar     foo  baz   no
2  foobar  barfoo  baz  yes

如果需要自定义功能，较大的DataFrame：

应该更慢

def somefunction (row):
    if 'foo' in row['a'] and row['b'] == 'bar':
        return 'yes'
    return 'no'

print (df.apply(somefunction, axis=1))
0    yes
1     no
2     no
dtype: object

def somefunction (row):
    if 'foo' in row['a']  and  'bar' in row['b']:
        return 'yes'
    return 'no'

print (df.apply(somefunction, axis=1))
0    yes
1     no
2    yes
dtype: object

<强>计时：

df = pd.concat([df]*1000).reset_index(drop=True)

def somefunction (row):
    if 'foo' in row['a'] and row['b'] == 'bar':
        return 'yes'
    return 'no'

In [269]: %timeit df['new'] = df.apply(somefunction, axis=1)
10 loops, best of 3: 60.7 ms per loop

In [270]: %timeit df['new1'] = np.where(df['a'].str.contains('foo') & (df['b'] == 'bar'), 'yes', 'no')
100 loops, best of 3: 3.25 ms per loop

df = pd.concat([df]*10000).reset_index(drop=True)

def somefunction (row):
    if 'foo' in row['a'] and row['b'] == 'bar':
        return 'yes'
    return 'no'

In [272]: %timeit df['new'] = df.apply(somefunction, axis=1)
1 loop, best of 3: 614 ms per loop

In [273]: %timeit df['new1'] = np.where(df['a'].str.contains('foo') & (df['b'] == 'bar'), 'yes', 'no')
10 loops, best of 3: 23.5 ms per loop

Answer 2

你的例外可能来自你写的事实

if row['a'].str.contains('foo')==True

删除＆＃39; .str＆＃39;：

if row['a'].contains('foo')==True

Python函数部分字符串匹配

2 个答案: