基于先前行的新列的Python条件值
基于先前行的新列的Python条件值
提问于 2020-12-04 14:38:15
为dataframe创建了5P移动平均值和20P移动平均值，我正在尝试基于5P和20P移动平均值的值创建一个新列。
如果5P移动平均值>20P移动平均值，则新的列单元格=‘买入’如果5P移动平均值<20P移动平均值，则新的列单元格=‘卖出’
我已经创建了一个专栏，上面写着买/卖
我正在尝试创建一个新列，它确认如果行和row+1不同，则根据订单买入/卖出。
这就是我的数据帧的样子
CBA = {5MA: [13.11,13.44,13.56,13.45,13.10,12.45],
20MA: [12.67,12.77,12.87,13.50,13.45,12.30],
test: [BUY,BUY,BUY,SELL,SELL,BUY]}我想要的是结果
CBA = {5MA: [13.11,13.44,13.56,13.45,13.10,12.45],
20MA: [12.67,12.77,12.87,13.50,13.45,12.30],
test: [BUY,BUY,BUY,SELL,SELL,BUY],
BUY/SELL: [NaN, NaN,SELL,NaN,BUY]}我制作了这个函数
def buy_sell_strat(df):
for i, row in df:
if df.loc[i,'test'] == 'SELL' & df.loc[i+1,'test'] == 'BUY':
return 'BUY'
elif df.loc[i,'test'] == 'BUY' & df.loc[i+1,'test'] == 'SELL':
return 'SELL'
else:
return 'NaN'
CBA['Buy/Sell'] = CBA.apply(buy_sell_strat,axis = 1)然而，我得到了错误
TypeError:无法解包不可迭代的浮点对象
回答 1
Stack Overflow用户
回答已采纳
发布于 2020-12-04 14:52:56
循环解决方案
为什么不直接使用(下面更快的解决方案)：
data = {"5MA": [13.11, 13.44, 13.56, 13.45, 13.10, 12.45],
"20MA": [12.67, 12.77, 12.87, 13.50, 13.45, 12.30],
"test": ["BUY", "BUY", "BUY", "SELL", "SELL", "BUY"]}
df = pd.DataFrame(data)
# takes 1.41 s ± 14.7 ms per loop (mean ± std. dev. of 7 runs, 3 loops each)
for i in range(df.index[-1]):
if df.loc[i, 'test'] == 'SELL' and df.loc[i+1, 'test'] == 'BUY':
df.loc[i, 'Buy/Sell'] = 'BUY'
elif df.loc[i, 'test'] == 'BUY' and df.loc[i+1, 'test'] == 'SELL':
df.loc[i, 'Buy/Sell'] = 'SELL'
else:
df.loc[i, 'Buy/Sell'] = 'NaN'输出为：
| | 5MA | 20MA | test | Buy/Sell |
|---:|------:|-------:|:-------|:-----------|
| 0 | 13.11 | 12.67 | BUY | NaN |
| 1 | 13.44 | 12.77 | BUY | NaN |
| 2 | 13.56 | 12.87 | BUY | SELL |
| 3 | 13.45 | 13.5 | SELL | NaN |
| 4 | 13.1 | 13.45 | SELL | BUY |
| 5 | 12.45 | 12.3 | BUY | nan |快速解决方案
多亏了@CainãMax Couto- pandas的np.nan提示，一种更快、更像席尔瓦的解决方案会更快(比6000行dataframe快500倍左右)。
df['Buy/Sell'] = np.nan
# takes 3.2 ms ± 52.1 µs per loop(mean ± std. dev. of 7 runs, 100 loops each)
df.loc[(df['test'] == "SELL") & (df['test'].shift(-1) == "BUY"), 'Buy/Sell'] = 'BUY'
df.loc[(df['test'] == "BUY") & (df['test'].shift(-1) == "SELL"), 'Buy/Sell'] = 'SELL'这给出了相同的输出：
| | 5MA | 20MA | test | Buy/Sell |
|---:|------:|-------:|:-------|:-----------|
| 0 | 13.11 | 12.67 | BUY | NaN |
| 1 | 13.44 | 12.77 | BUY | NaN |
| 2 | 13.56 | 12.87 | BUY | SELL |
| 3 | 13.45 | 13.5 | SELL | NaN |
| 4 | 13.1 | 13.45 | SELL | BUY |
| 5 | 12.45 | 12.3 | BUY | NaN |页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/65139011
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