如果在'Gap Fill?'列中有'1'，我想在'Gap Down?'列中放入1，并且从今天开始一直到'Highs'的最大值未来5天，或者(.shift(-5))是昨天的>=或'Adj Close'的{{1}}。
这是我的代码，可以重新创建并提出解决方案：
(.shift(-1))这当前提供输出：
# 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
最后，我想这样：
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
...
这是为了检查是否存在差距，以及是否在接下来的5天之内填补了差距。谢谢！
答案 0 :(得分：1)
您可以这样做：
# 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，对于满足条件1且其索引存储在ind中的所有元素，条件1被检查，条件2被循环检查。在这种情况下，只有一个，但是当有多个时，您可以这样做。
我想我了解您的要求，如果第二个条件不完全符合您的要求，您可以对其进行检查并调整必要的值。