我有一个循环,它创建一个更大数据帧的子集,然后将这些子集用作函数的输入进行分析。此函数返回一个列表。我用指纹看它停在哪里。因此,对于循环中的第一次运行,我可以看到子集,函数的列表输出,但是当它再次启动时,第二次运行,我看到第二个子集,但是在函数处,我得到以下错误:
<ipython-input-11-8f6203e297e3> in ssd(x, y)
8
9 for i in range(x.shape[0]):
---> 10 spread_cumdiff += (x[i] - y[i]) **2
11
12 return spread_cumdiff
注意上面的部分是python错误之前的最后一部分。实际上,上面还有两个类似的块,即函数a)包含函数b),函数b包含上面的块。你知道吗
~/anaconda3/envs/thesis/lib/python3.5/site-packages/pandas/core/series.py in __getitem__(self, key)
621 key = com._apply_if_callable(key, self)
622 try:
--> 623 result = self.index.get_value(self, key)
624
625 if not is_scalar(result):
~/anaconda3/envs/thesis/lib/python3.5/site-packages/pandas/core/indexes/base.py in get_value(self, series, key)
2558 try:
2559 return self._engine.get_value(s, k,
-> 2560 tz=getattr(series.dtype, 'tz', None))
2561 except KeyError as e1:
2562 if len(self) > 0 and self.inferred_type in ['integer', 'boolean']:
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_value()
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_value()
pandas/_libs/index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.Int64HashTable.get_item()
pandas/_libs/hashtable_class_helper.pxi in pandas._libs.hashtable.Int64HashTable.get_item()
KeyError: 0
我有这样的想法:
df_test = pd.DataFrame(np.random.randint(low=0, high=10, size=(5, 5)),
columns=['a', 'b', 'c', 'd', 'e'],
index = ['20100101', '20100102', '20100103', '20100104', '20100105']
dfs = []
N = 3
for x in np.arange(len(df_test)+1)[N:]:
df1 = df_test.iloc[np.arange(x - N, x)]
test_list = myfunc(df1) # it takes in df1, makes some computation and returns a
# list of 2-element tuples, i.e. [('a', 'b'), ('d', 'e')]
编辑:请参见下面的功能:
def ssd(x, y):
spread_cumdiff = 0
for i in range(x.shape[0]):
#print("x_i", x[i])
#print("y_i", y[i])
spread_cumdiff += (x[i] - y[i]) **2
return spread_cumdiff
我试着使用print函数,但在第二次运行循环时,它甚至都没有实现。你知道吗
def pairs_match(df, p):
df_norm = df.assign(**df.drop('datetime', 1).pipe(lambda d: d.div(d.shift().bfill()).cumprod()))
df_norm = df_norm.replace([np.inf, -np.inf], np.nan)
df_norm.fillna(method = 'ffill', inplace = True)
df_norm.fillna(method = 'bfill', inplace = True)
ticker = df_norm.columns.values.tolist()
ticker.pop(0)
ticker_list = pd.DataFrame({'ticker': ticker})
# to be implemented: if length of list list <2, then skip the entire run!
all_pairs = list(itertools.permutations(ticker_list.ticker, 2))
squared = []
presel_pairs = []
for i in all_pairs:
squared.append(ssd(df_norm[i[0]].head(n = train_win), df_norm[i[1]].head(n = train_win))) # ssd(x,y) function from above
tbl_dist = pd.DataFrame({'Pair' : all_pairs, 'SSD' : squared})
ssd_perctl = p
ssd_thresh = stats.scoreatpercentile(tbl_dist['SSD'], ssd_perctl)
presel_pairs = tbl_dist[tbl_dist['SSD'] <= ssd_thresh]
presel_pairs_list = presel_pairs['Pair']
presel_pairs_list = presel_pairs_list.reset_index(drop = True)
return presel_pairs_list
def pairs_match(df, p)
返回一个列表,然后在另一个函数中使用。你知道吗
试着分别打印
x[i]
和y[i]
,这样就可以知道这两个选项中的哪一个会导致keyerror。也请张贴的功能,因为没有它,我们不知道发生了什么。你知道吗正如我们所想,问题确实出在cumsum update函数上。我将函数重写如下:
这并不是问题的根源,但是它避免了循环。你知道吗
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