python - 使用 Pandas 检查股票价格的差距？
问题描述
'Gap Fill?'如果列中有a ，我想在列中输入1，'1'并且从今天开始到未来5天'Gap Down?'的最大值，或者是昨天的或。'Highs'(.shift(-5))>='Adj Close'(.shift(-1))
这是我重新创建并提出解决方案的代码：
# Import pandas library
import pandas as pd
# initialize list of lists
data = [[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[3, 3, 3, 3], [3, 3, 3, 3], [3, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4]
]
# Create the pandas DataFrame
df = pd.DataFrame(data, columns = ['Open', 'High', 'Low', 'Adj Close'])
# Now apply a new column and put a 1 if today's opened is less than yesterday's Adj Close
df['Gap Down?'] = 0
df.loc[df['Open'] < df['Adj Close'].shift(1), 'Gap Down?'] = 1
df.loc[df['Open'] >= df['Adj Close'].shift(1), 'Gap Down?'] = 0
# print the new dataframe
print(df)
# Now make a new column that puts a 1 if the highest price in the High column from today
# and FORWARD (future) 5 days becomes >= yesterday's close
df['Gap Fill?'] = 0
这当前给出了输出：
Open High Low Adj Close Gap Down?
0 5 10 3 4 0
1 5 10 3 4 0
2 5 10 3 4 0
3 5 10 3 4 0
4 5 10 3 4 0
5 5 10 3 4 0
6 5 10 3 4 0
7 5 10 3 4 0
8 5 10 3 4 0 <- Yesterday's Adj Close was 4
9 3 3 3 3 1 <- Today opened at 3 so there's been a gap down
10 3 3 3 3 0
11 3 10 3 4 0 <- Here is when the High went >= 4, and it's within 5 days of the open price in question
12 5 10 3 4 0
13 5 10 3 4 0
14 5 10 3 4 0
15 5 10 3 4 0
16 5 10 3 4 0
17 5 10 3 4 0
18 5 10 3 4 0
19 5 10 3 4 0
20 5 10 3 4 0
21 5 10 3 4 0
22 5 10 3 4 0
23 5 10 3 4 0
我想结束这个：
Open High Low Adj Close Gap Down? Gap Fill?
0 5 10 3 4 0 0
1 5 10 3 4 0 0
2 5 10 3 4 0 0
3 5 10 3 4 0 0
4 5 10 3 4 0 0
5 5 10 3 4 0 0
6 5 10 3 4 0 0
7 5 10 3 4 0 0
8 5 10 3 4 0 0
9 3 3 3 3 1 1 <- ... but the 1 should go here
10 3 3 3 3 0 0
11 3 10 3 4 0 0 <- The gap fill happened here two days later...
12 5 10 3 4 0 0
13 5 10 3 4 0 0
14 5 10 3 4 0 0
15 5 10 3 4 0 0
16 5 10 3 4 0 0
17 5 10 3 4 0 0
18 5 10 3 4 0 0
19 5 10 3 4 0 0
...
这是为了检查是否存在缺口，以及是否在接下来的 5 天中的某个时间点被填补。谢谢！
解决方案
你可以这样做：
# Import pandas library
import pandas as pd
# initialize list of lists
data = [[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[3, 3, 3, 3], [3, 3, 3, 3], [3, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4],
[5, 10, 3, 4], [5, 10, 3, 4], [5, 10, 3, 4]
]
# Create the pandas DataFrame
df = pd.DataFrame(data, columns = ['Open', 'High', 'Low', 'Adj Close'])
# Now apply a new column and put a 1 if today's opened is less than yesterday's Adj Close
df['Gap Down?'] = 0
df.loc[df['Open'] < df['Adj Close'].shift(1), 'Gap Down?'] = 1
df.loc[df['Open'] >= df['Adj Close'].shift(1), 'Gap Down?'] = 0
# print the new dataframe
print(df)
# Now make a new column that puts a 1 if the highest price in the High column from today
# and FORWARD (future) 5 days becomes >= yesterday's close
df['Gap Fill?'] = df['Gap Down?']
#first condition
e=df['Gap Down?'].eq(1)
ind=df[e].index.values.tolist()
i=0
while i<len(ind):
#cheked of second condition
if(df.loc[ind[i]:ind[i]+4,'High'].max()>=df.loc[ind[i-1],'Adj Close']):
df.loc[ind[i],'Gap Fill?'] = 1
i+=1
print(df)
输出：
Open High Low Adj Close Gap Down?
0 5 10 3 4 0
1 5 10 3 4 0
2 5 10 3 4 0
3 5 10 3 4 0
4 5 10 3 4 0
5 5 10 3 4 0
6 5 10 3 4 0
7 5 10 3 4 0
8 5 10 3 4 0
9 3 3 3 3 1
10 3 3 3 3 0
11 3 10 3 4 0
12 5 10 3 4 0
13 5 10 3 4 0
14 5 10 3 4 0
15 5 10 3 4 0
16 5 10 3 4 0
17 5 10 3 4 0
18 5 10 3 4 0
19 5 10 3 4 0
20 5 10 3 4 0
21 5 10 3 4 0
22 5 10 3 4 0
23 5 10 3 4 0
Open High Low Adj Close Gap Down? Gap Fill?
0 5 10 3 4 0 0
1 5 10 3 4 0 0
2 5 10 3 4 0 0
3 5 10 3 4 0 0
4 5 10 3 4 0 0
5 5 10 3 4 0 0
6 5 10 3 4 0 0
7 5 10 3 4 0 0
8 5 10 3 4 0 0
9 3 3 3 3 1 1
10 3 3 3 3 0 0
11 3 10 3 4 0 0
12 5 10 3 4 0 0
13 5 10 3 4 0 0
14 5 10 3 4 0 0
15 5 10 3 4 0 0
16 5 10 3 4 0 0
17 5 10 3 4 0 0
18 5 10 3 4 0 0
19 5 10 3 4 0 0
20 5 10 3 4 0 0
21 5 10 3 4 0 0
22 5 10 3 4 0 0
23 5 10 3 4 0 0
使用e，条件 1 被检查，条件 2 由循环检查所有满足条件 1 且索引存储在 中的元素ind。在这种情况下，只有一个，但这允许您在有多个时执行此操作。
我想我明白你在找什么，如果第二个条件不是你所要求的，你可以检查它并调整必要的值。
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