列表中数值，怎么放入函数中一一判断呢？

由iquant创建，最终由iquant更新于2023-06-01 02:13 被浏览 47 用户

问题

老师，麻烦您看下；第一段，是选出一系列股票 xuangu=【…】第二段，是对某个股票做出判断，最终正确，就输出 “true”，我把这个判断，做成了一个函数第三段，我怎么把第一段选出的股票，一个个放入 instruments(a)里的a中，如果判断正确，就输出股票

l=DataSource(‘bar1d_CN_STOCK_A’).read(instruments=[],start_date=‘2021-06-10’,end_date=None,fields=[‘open’,‘close’,‘high’,‘low’,‘amount’,‘adjust_factor’])
l[‘open+’]=l[‘open’]/l[‘adjust_factor’]
l[‘close+’]=l[‘close’]/l[‘adjust_factor’]
l[‘high+’]=l[‘high’]/l[‘adjust_factor’]
l[‘low+’]=l[‘low’]/l[‘adjust_factor’]
c1 = l[‘date’]>=‘2021-06-08’
c2 = l[‘amount’]>100000000
c3 = l[‘close+’]>4.99
l1 = l.loc[c1&c2&c3,[‘instrument’,‘date’,‘open+’,‘high+’,‘close+’,‘low’,‘amount’]]
xuangu= l1[‘instrument’].values.tolist()
def instruments(a):
instruments=a
df = DataSource(“bar60m_CN_STOCK_A”).read(a,start_date=“2020-03-10”)
df = df[[‘open’,‘high’,‘close’,‘low’]]
data = df.values.tolist()
data.reverse()

a=data[0][1]
i=0
for i in range(len(data)):
if(data[i][1] > a):
break
b=i
m = data[0:b+1]
result=[]
for i in range(0,len(m)):
result=list(set(result).union(set(m[i])))
one = min(result)

n=data[b:]
for i in range(len(n)):
if(n[i][3] <=one):
break
c=i
o=n[0:c+1]
result=[]
for i in range(0,len(o)):
result=list(set(result).union(set(o[i])))
two = max(result)

p=n[c:]
for i in range(len(p)):
if(p[i][1] >=two):
break
d=i
q=p[0:i+1]
result=[]
for i in range(0,len(q)):
result=list(set(result).union(set(q[i])))three = min(result)

r=p[d:]
for i in range(len(r)):
if(r[i][3] <=three):
break
e=i
s=r[0:i+1]
result=[]
for i in range(0,len(s)):
result=list(set(result).union(set(s[i])))four = max(result)
five=min(result)

print(one,two,three,four,five)

if(four-three)*0.585<(two-one) and b>0 and c>0 and d>0 and e>0 and b+c+d+e<1000:
print("True")
else:print("Flase")

条件 c1 = l[‘date’]>='2021-06-08’是多余的，因为你的start_date已经是2021-06-10
不建议在循环里面调用DataSource，效率低，可以参照如下代码，先读取然后再通过股票来过滤
下次分享代码请点击策略编辑页面右上角的策略连接分享按钮，把链接发出来就可以了，不然直接这样拷贝代码格式会乱

{w:100}

def test(instr,df60):
df1 = df60[df60['instrument']==instr]
df = df1[["open","high","close","low"]]
data = df.values.tolist()
data.reverse()
a=data[0][1]
i=0
for i in range(len(data)):
if(data[i][1] > a):
break
b=i
m = data[0:b+1]
result=[]
for i in range(0,len(m)):
result=list(set(result).union(set(m[i])))
one = min(result)
n=data[b:]
for i in range(len(n)):
    if(n[i][3] <=one):
        break
c=i
o=n[0:c+1]
result=[]
for i in range(0,len(o)):
    result=list(set(result).union(set(o[i])))
two = max(result)

p=n[c:]
for i in range(len(p)):
    if(p[i][1] >=two):
        break
d=i
q=p[0:i+1]
result=[]
for i in range(0,len(q)):
    result=list(set(result).union(set(q[i])))  
three = min(result)


r=p[d:]
for i in range(len(r)):
    if(r[i][3] <=three):
        break
e=i
s=r[0:i+1]
result=[]
for i in range(0,len(s)):
    result=list(set(result).union(set(s[i])))  
four = max(result)
five=min(result)

print(one,two,three,four,five)

if(four-three)*0.585<(two-one) and b>0 and c>0 and d>0 and e>0 and b+c+d+e<1000:
    print("True")
else:   
    print("Flase")
l=DataSource('bar1d_CN_STOCK_A').read(instruments=[],start_date='2021-06-10',end_date=None,fields=['open','close','high','low','amount','adjust_factor'])
l['open+']=l['open']/l['adjust_factor']
l['close+']=l['close']/l['adjust_factor']
l['high+']=l['high']/l['adjust_factor']
l['low+']=l['low']/l['adjust_factor']
c1 = l['date']>='2021-06-08'
c2 = l['amount']>100000000
c3 = l['close+']>4.99
l1 = l.loc[c1&c2&c3,['instrument','date','open+','high+','close+','low','amount']]
xuangu= l1['instrument'].values.tolist()

df60 = DataSource("bar60m_CN_STOCK_A").read(instruments=xuangu,start_date="2021-03-10")
print(df60)
for instr in xuangu:
test(instr,df60)

列表中数值，怎么放入函数中 一一判断呢？

问题

标签

列表中数值，怎么放入函数中一一判断呢？