我正在尝试使用apply和lambda函数替换下面注释掉的for循环，但我收到以下错误。我的python是生锈的，所以任何提示都非常感激。
错误：
File "<ipython-input-5-b29bfb93595e>", line 11
if (dataDF < dataDF.shift()) & (dataDF.shift(periods=1) < dataDF.shift(periods=2)):
^
SyntaxError：语法无效
代码：
def get_recession_end():
dataDF = pd.ExcelFile('gdplev.xls').parse(skiprows=7)[['Unnamed: 4', 'Unnamed: 5']].loc[246:]
dataDF.columns = ['Quarter','dataDF']
dataDF['dataDF'] = pd.to_numeric(dataDF['dataDF'])
#quarters = []
#for i in range(len(dataDF) - 2):
#if (dataDF.iloc[i][1] < dataDF.iloc[i+1][1]) & (dataDF.iloc[i+1][1] < dataDF.iloc[i+2][1]):
#quarters.append(dataDF.iloc[i+2][0])
#return quarters[0]
quarters = dataDF.apply(lambda x: quarters = []
if (dataDF < dataDF.shift()) & (dataDF.shift(periods=1) < dataDF.shift(periods=2)):
quarters.append(dataDF.shift(2)[0]))
return quarters[0]
get_recession_end()
更新新版本：
代码：
def get_recession_end():
def get_recession_end():
dataDF = pd.ExcelFile('gdplev.xls').parse(skiprows=7)[['Unnamed: 4',
'Unnamed: 5']].loc[246:]#skiprows=17,skip_footer=(38))
dataDF.columns = ['Quarter','dataDF']
dataDF['dataDF'] = pd.to_numeric(dataDF['dataDF'])
#quarters = []
#for i in range(len(dataDF) - 2):
#if (dataDF.iloc[i][1] < dataDF.iloc[i+1][1]) & (dataDF.iloc[i+1][1]
< dataDF.iloc[i+2][1]):
#quarters.append(dataDF.iloc[i+2][0])
#return quarters[0]
def do_the_foo(x):
quarters = []
if (dataDF < dataDF.shift()) & (dataDF.shift(periods=1) <
dataDF.shift(periods=2)):
quarters.append(dataDF.shift(2)[0])
return quarters
quarters = dataDF.loc[:(len(dataDF) - 2)].apply(do_the_foo)
return quarters[0]
get_recession_end()
新错误：
---------------------------------------------------------------------------
KeyError Traceback (most recent call last)
/opt/conda/lib/python3.5/site-packages/pandas/indexes/base.py in
get_loc(self, key, method, tolerance)
1944 try:
-> 1945 return self._engine.get_loc(key)
1946 except KeyError:
pandas/index.pyx in pandas.index.IndexEngine.get_loc (pandas/index.c:4154)()
pandas/index.pyx in pandas.index.IndexEngine.get_loc (pandas/index.c:4018)()
pandas/hashtable.pyx in pandas.hashtable.PyObjectHashTable.get_item
(pandas/hashtable.c:12368)()
pandas/hashtable.pyx in pandas.hashtable.PyObjectHashTable.get_item
(pandas/hashtable.c:12322)()
KeyError: 0
During handling of the above exception, another exception occurred:
KeyError Traceback (most recent call last)
<ipython-input-10-53e0a21f9faa> in <module>()
18
19
---> 20 get_recession_end()
<ipython-input-10-53e0a21f9faa> in get_recession_end()
15
16 quarters = dataDF.loc[:-(len(dataDF) - 2)].apply(do_the_foo)
---> 17 return quarters[0]
18
19
/opt/conda/lib/python3.5/site-packages/pandas/core/frame.py in
__getitem__(self, key)
1995 return self._getitem_multilevel(key)
1996 else:
-> 1997 return self._getitem_column(key)
1998
1999 def _getitem_column(self, key):
/opt/conda/lib/python3.5/site-packages/pandas/core/frame.py in
_getitem_column(self, key)
2002 # get column
2003 if self.columns.is_unique:
-> 2004 return self._get_item_cache(key)
2005
2006 # duplicate columns & possible reduce dimensionality
/opt/conda/lib/python3.5/site-packages/pandas/core/generic.py in
_get_item_cache(self, item)
1348 res = cache.get(item)
1349 if res is None:
-> 1350 values = self._data.get(item)
1351 res = self._box_item_values(item, values)
1352 cache[item] = res
/opt/conda/lib/python3.5/site-packages/pandas/core/internals.py in get(self,
项目，快速路径） 3288 3289如果不是isnull（item）： - ＆GT; 3290 loc = self.items.get_loc（item） 3291其他： 3292 indexer = np.arange（len（self.items）） [ISNULL（self.items）]
/opt/conda/lib/python3.5/site-packages/pandas/indexes/base.py in
get_loc(self, key, method, tolerance)
1945 return self._engine.get_loc(key)
1946 except KeyError:
-> 1947 return
self._engine.get_loc(self._maybe_cast_indexer(key))
1948
1949 indexer = self.get_indexer([key], method=method,
tolerance=tolerance)
pandas/index.pyx in pandas.index.IndexEngine.get_loc (pandas/index.c:4154)()
pandas/index.pyx in pandas.index.IndexEngine.get_loc (pandas/index.c:4018)()
pandas/hashtable.pyx in pandas.hashtable.PyObjectHashTable.get_item
(pandas/hashtable.c:12368)()
pandas/hashtable.pyx in pandas.hashtable.PyObjectHashTable.get_item
(pandas/hashtable.c:12322)()
KeyError: 0
答案 0 :(得分：0)
lambda expressions仅限于一个表达式。您已尝试使用多个语句。 lambda表达式只是内联一个小函数的一种方便方法，并且总是可以被常规函数替换。由于lambda引用了外部函数作用域中的变量，因此还应该在该作用域中定义替换函数。
结果看起来并不比你想要替换的原件好。所以，我认为真正的答案是你不能这样使用lambda。
def get_recession_end():
dataDF = pd.ExcelFile('gdplev.xls').parse(skiprows=7)[['Unnamed: 4', 'Unnamed: 5']].loc[246:]
dataDF.columns = ['Quarter','dataDF']
dataDF['dataDF'] = pd.to_numeric(dataDF['dataDF'])
#quarters = []
#for i in range(len(dataDF) - 2):
#if (dataDF.iloc[i][1] < dataDF.iloc[i+1][1]) & (dataDF.iloc[i+1][1] < dataDF.iloc[i+2][1]):
#quarters.append(dataDF.iloc[i+2][0])
#return quarters[0]
def do_the_foo(x):
quarters = []
if (dataDF < dataDF.shift()) & (dataDF.shift(periods=1) < dataDF.shift(periods=2)):
quarters.append(dataDF.shift(2)[0])
return quarters
quarters = dataDF.apply(do_the_foo)
get_recession_end()