[英]Index of the previous value that is greater than or equal 1.2 times the current value
對於任何給定日期,我試圖找到比當前close
價高 1.2 倍的前一個close
價。 我做了一個循環來檢查每一行。 但是,它效率不高。 運行時間為 45 秒。 如何使我的代碼更有效地處理比這大得多的數據集?
數據集 - TSLA或TSLA Daily 5Y Stock Yahoo
df = pd.read_csv(os.getcwd()+"\\TSLA.csv")
# Slicing the dataset
df2 = df[['Date', 'Close']]
irange = np.arange(1, len(df))
for i in irange:
# Dicing first i rows
df3 = df2.head(i)
# Set the target close value that is 1.2x the current close value
targetValue = 1.2 * df3['Close'].tail(1).values[0]
# Check the last 200 days
df4 = df3.tail(200)
df4.set_index('Date', inplace=True)
# Save all the target values in a list
req = df4[df4['Close'] > targetValue]
try:
lent = (req.index.tolist()[-1])
except:
lent = str(9999999)
# Save the last value to the main dataframe
df.at[i,'last_time'] = lent
df.tail(20)
你正在做 O(N^3) 和一些不必要的數據副本。 試試這個 O(NlogN) 方式
df = pd.read_csv("D:\\TSLA.csv")
stack,cnt=[],0
def OnePointTwoTimesLarger(row):
#cnt is not really needed by what you aksed. But it is usually a better to return the data row you need, instead of just returning the value
global stack,cnt
c=row['Close']
while stack and stack[-1][1]<=c:
stack.pop()
stack.append([row['Date'],c])
cnt+=1
left,right=0,len(stack)-1
while left<right-3:
mid=(left+right)//2
if stack[mid][1]>1.2*c:
left=mid
else:
right=mid
for e in stack[left:right+1][::-1]:
if e[1]>1.2*c:
return e[0]
return 999999
df['last_time']=df.apply(OnePointTwoTimesLarger, axis=1)
df.tail(60)
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