遍历股票价格的行和列python

Question

我下面有一个代码

result, diff = [], []

for index, row in final.iterrows():
    for column in final.columns:
        if ((final['close'] - final['open']) > 20):
            diff = final['close'] - final['open']
            result = 1
        elif ((final['close'] - final['open']) < -20):
            diff = final['close'] - final['open']
            result = -1
        elif (-20 < (final['close'] - final['open']) < 20 ):
            diff = final['close'] - final['open']
            result = 0
        else:
            continue

目的是针对每个时间戳，检查收盘价-开盘价是否大于20点，然后为其分配买入价。 如果小于-20，则分配一个卖出价格；如果介于两者之间，则分配一个0。

我收到此错误

The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
[Finished in 35.418s

Answer 1

曾经对pandas有经验的人会给出更好的答案，但由于没人在这里回答我的问题。 您通常不希望直接使用pandas.Dataframes进行迭代，因为这样做会达到目的。 pandas解决方案看起来更像是：

import pandas as pd

data = {
    'symbol': ['WZO', 'FDL', 'KXW', 'GYU', 'MIR', 'YAC', 'CDE', 'DSD', 'PAM', 'BQE'],
    'open': [356, 467, 462, 289, 507, 654, 568, 646, 440, 625],
    'close': [399, 497, 434, 345, 503, 665, 559, 702, 488, 608]
}
df = pd.DataFrame.from_dict(data)
df['diff'] = df['close'] - df['open']
df.loc[(df['diff'] < 20) & (df['diff'] > -20), 'result'] = 0
df.loc[df['diff'] >= 20, 'result'] = 1
df.loc[df['diff'] <= -20, 'result'] = -1

df现在包含：

  symbol  open  close  diff  result
0    WZO   356    399    43     1.0
1    FDL   467    497    30     1.0
2    KXW   462    434   -28    -1.0
3    GYU   289    345    56     1.0
4    MIR   507    503    -4     0.0
5    YAC   654    665    11     0.0
6    CDE   568    559    -9     0.0
7    DSD   646    702    56     1.0
8    PAM   440    488    48     1.0
9    BQE   625    608   -17     0.0

关于您的代码，我将从上面重复我的评论：您正在按row进行迭代，然后在您的条件下使用整个DataFrame final 。 我想你打算在那里row 。 您无需遍历按索引获取值的列。 当final['close'] - final['open']正好为20时，您的条件会丢失。 result, diff = [], []是顶部列表，但随后在循环中分配为整数。 也许您想要result.append() ？