Question

我正在尝试填充看起来像这样的数据框

      Name   Origin      Date Open High  Low Close    Date+1  Open+1 High+1 Low+1 Close+1
0  Bananas     Bali  20200108  NaN  NaN  NaN   NaN  20200109     NaN    NaN   NaN     NaN
1  Coconut  Bahamas  20200110  NaN  NaN  NaN   NaN  20200111     NaN    NaN   NaN     NaN

在看起来像这样的数据框中找到数据

      Name   Origin      Date      Time  Open  High  Low  Close
0  Bananas     Bali  20200108  15:30:00  1.58  1.85  1.4   1.50
1  Bananas     Bali  20200108  22:00:00  1.68  1.78  1.5   1.60
2  Bananas     Bali  20200109  15:30:00  1.88  1.95  1.7   1.86
3  Bananas     Bali  20200109  22:00:00  1.78  1.88  1.6   1.65
4  Coconut  Bahamas  20200110  15:30:00  2.58  2.85  2.4   2.50
5  Coconut  Bahamas  20200110  22:00:00  2.68  2.78  2.5   2.60
6  Coconut  Bahamas  20200111  15:30:00  2.88  2.95  2.7   2.86
7  Coconut  Bahamas  20200111  22:00:00  2.78  2.88  2.6   2.65

由于第一个数据框中的列具有不同的名称（例如，“ Open”和“ Open + 1”），因此我想不出一种简单的方法来进行索引匹配，而不必复制代码并重命名列在第二个数据帧中。因此，我认为按列号索引匹配更容易，但是即时通讯在确定如何执行此操作方面存在问题。列的条件为“名称”，“来源”和“日期”（Open + 1的日期+1，等等）。

我尝试使用以下代码：

ColOpen = df2.iloc[:, [0,1,2,4,5,6,7]].groupby([0,1,2]).agg(Open=(4,'first'),High=(5,'max'),Low=(6,'min'), Close=(7,'last'))

获得正确的列值，但是我得到的是'KeyError：0'，它引用列号。

我在下面创建了一个示例代码，可用于获取相同的数据帧。

import pandas as pd

#Creating first sample dataframe
lst1 = [['Bananas', 'Bali', '20200108', 'NaN', 'NaN', 'NaN', 'NaN', '20200109', 'NaN', 'NaN', 'NaN', 'NaN'],
   ['Coconut', 'Bahamas', '20200110', 'NaN', 'NaN', 'NaN', 'NaN', '20200111', 'NaN', 'NaN', 'NaN', 'Nan']]

df1 = pd.DataFrame(lst1, columns =['Name', 'Origin', 'Date', 'Open', 'High', 'Low', 'Close', 'Date+1', 'Open+1', 'High+1', 'Low+1', 'Close+1'])
print('First Dataframe')
print(df1)

#Creating second sample dataframe
lst2 = [['Bananas', 'Bali', '20200108', '15:30:00', 1.58, 1.85, 1.50, 1.50],
    ['Bananas', 'Bali', '20200108', '22:00:00', 1.68, 1.78, 1.40, 1.60],
    ['Bananas', 'Bali', '20200109', '15:30:00', 1.88, 1.95, 1.70, 1.86],
    ['Bananas', 'Bali', '20200109', '22:00:00', 1.78, 1.88, 1.60, 1.65],
    ['Coconut', 'Bahamas', '20200110', '15:30:00', 2.58, 2.85, 2.50, 2.50],
    ['Coconut', 'Bahamas', '20200110', '22:00:00', 2.68, 2.78, 2.40, 2.60],
    ['Coconut', 'Bahamas', '20200111', '15:30:00', 2.88, 2.95, 2.70, 2.86],
    ['Coconut', 'Bahamas', '20200111', '22:00:00', 2.78, 2.88, 2.60, 2.65]]

df2 = pd.DataFrame(lst2, columns =['Name', 'Origin', 'Date', 'Time', 'Open', 'High', 'Low', 'Close'])
print('Second Dataframe')
print(df2)

#Index Match

ColOpen = df2.iloc[:, [0,1,2,4,5,6,7]].groupby([0,1,2]).agg(Open=(4,'first'),High=(5,'max'),Low=(6,'min'), Close=(7,'last'))


print("Printing first index")
print(ColOpen)

#Desired Output
lst3 = [['Bananas', 'Bali', '20200108', 1.58, 1.85, 1.4, 1.6, '20200109', 1.88, 1.95, 1.6, 1.65],
   ['Coconut', 'Bahamas', '20200110', 2.58, 2.85, 2.4, 2.6, '20200111', 2.88, 2.95, 2.6, 2.65]]

df3 = pd.DataFrame(lst3, columns =['Name', 'Origin', 'Date', 'Open', 'High', 'Low', 'Close', 'Date+1', 'Open+1', 'High+1', 'Low+1', 'Close+1'])
print('Desired Output')
print(df3)

有人可以帮我弄清楚该怎么做吗？

编辑：所需的输出。还更新了一些代码。

      Name   Origin      Date  Open  ...  Open+1  High+1  Low+1 Close+1
0  Bananas     Bali  20200108  1.58  ...    1.88    1.95    1.6    1.65
1  Coconut  Bahamas  20200110  2.58  ...    2.88    2.95    2.6    2.65

Answer 1

编辑：使用groupby找到了更简单的解决方案。

基本上，您pd.concat是您的数据，但是您concat的数据是shift后退1行。然后concat，并进行一些编辑。你有它！ df4是您要寻找的。

import pandas as pd

df = pd.read_clipboard()

# all your new data is here
df2 = df.groupby(["Date", "Name", "Origin"]).agg(
    {"Open": ["min"], "High": ["max"], "Low": ["min"], "Close": ["max"]}
)

df2 = df2.droplevel(1, axis=1).reset_index()

column_names = ["Name", "Origin", "Date", "Open", "High", "Low", "Close", "Date+1", "Open+1", "High+1", "Low", "Close+1"]
desired_df = pd.DataFrame(columns=column_names)

df3 = pd.concat([df2, df2.add_suffix('+1').shift(-1)], axis=1)

df4 = df3.iloc[::2]

df4 = df4.drop(columns=['Date+1', 'Name+1', 'Origin+1']).reset_index(drop=True)

    Date    Name    Origin  Open    High    Low Close   Open+1  High+1  Low+1   Close+1
0   20200108    Bananas Bali    1.58    1.85    1.4 1.6 1.78    1.95    1.6 1.86
1   20200110    Coconut Bahamas 2.58    2.85    2.4 2.6 2.78    2.95    2.6 2.86

这不是最有效的答案，但是期望的结果是如此不寻常。这是代码，我主要将Python函数与pandas dataframes一起使用。通过使用Ctrl+C复制表或手动添加表来获取数据。

import pandas as pd
import numpy as np

df = pd.read_clipboard()
column_names = ["Name", "Origin", "Date", "Open", "High", "Low", "Close", "Date+1", "Open+1", "High+1", "Low", "Close+1"]

def data_getter(data):
    intro = data.iloc[0][0:3]
    open_ = data.iloc[0].Open
    close = data.iloc[1].Close
    high = data.loc[:, 'High'].max()
    low = data.loc[:, 'Low'].min()
    frame = np.append(intro, [open_, high, low, close])
    return frame

def df_formatter(num: int):

    d = []

    for i in range(2):
        data = df.iloc[num*4+(i)*2:num*4+(i+1)*2]
        d.append(data_getter(data))

    d = np.append(d[0], [d[1][2:]])
    d = pd.Series(d)
    d.index = column_names
    return d

desired_df = pd.DataFrame(columns=column_names)

for i in range(int(df.shape[0]/4)):
    desired_df = desired_df.append(df_formatter(i), ignore_index=True)

print(desired_df)

熊猫索引与列号和多个条件匹配

1 个答案: