遍歷股票價格的行和列python

Question

我下面有一個代碼

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

目的是針對每個時間戳，檢查收盤價-開盤價是否大於20點，然后為其分配買入價。 如果小於-20，則分配一個賣出價格；如果介於兩者之間，則分配一個0。

我收到此錯誤

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

曾經對pandas有經驗的人會給出更好的答案，但由於沒人在這里回答我的問題。 您通常不希望直接使用pandas.Dataframes進行迭代，因為這樣做會達到目的。 pandas解決方案看起來更像是：

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現在包含：

  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

關於您的代碼，我將從上面重復我的評論：您正在按row進行迭代，然后在您的條件下使用整個DataFrame final 。 我想你打算在那里row 。 您無需遍歷按索引獲取值的列。 當final['close'] - final['open']正好為20時，您的條件會丟失。 result, diff = [], []是頂部列表，但隨后在循環中分配為整數。 也許您想要result.append() ？