查找连续相似值的最大值

Question

我有一个像这样的数据框：

               timestamp  Order     Price  Quantity
0    2019-10-09 09:15:42      0  27850.00      2040
1    2019-10-09 09:15:42      0  27850.00      1980
2    2019-10-09 09:15:53      0  27860.85      1800
3    2019-10-09 09:16:54      0  27860.85      2340
4    2019-10-09 09:18:48      0  27860.85      1500
5    2019-10-09 09:21:08      0  27979.00      1840
6    2019-10-09 09:21:08      0  27979.00      2020
7    2019-10-09 09:21:12      0  27850.00      1800
8    2019-10-09 09:21:15      0  27850.00      1580
9    2019-10-09 09:21:21     35  28000.00      1840
10   2019-10-09 09:21:23     34  28000.00      1800
11   2019-10-09 09:28:17      0  28035.00      2020
12   2019-10-09 09:28:18      0  28035.00      1960
13   2019-10-09 09:28:18      0  28035.00      1920
14   2019-10-09 09:28:24      0  28035.00      1940
15   2019-10-09 09:28:24      0  28035.00      1960
16   2019-10-09 09:28:25      0  28000.00      2140
17   2019-10-09 09:28:25      0  28000.00      2020
18   2019-10-09 09:28:26      0  28000.00      2120

我想检查连续的价格值何时相同，然后返回具有最大数量值的行。

我的结果数据框喜欢：

               timestamp  Order     Price  Quantity
0    2019-10-09 09:15:42      0  27850.00      2040
3    2019-10-09 09:16:54      0  27860.85      2340
6    2019-10-09 09:21:08      0  27979.00      2020
7    2019-10-09 09:21:12      0  27850.00      1800
9    2019-10-09 09:21:21     35  28000.00      1840
11   2019-10-09 09:28:17      0  28035.00      2020
16   2019-10-09 09:28:25      0  28000.00      2140

PS：结果表中价格值27850.00再次出现在第7行，将被视为独立。 28000.00 也是如此。

Answer 1

首先创建一个price_group 列来识别具有相同价格的连续行（如this answer）。

price_group = (df.Price != df.Price.shift()).cumsum()

然后按此列对行进行分组，并找到每个组的最大数量的行（如theseanswers）。

result = df.loc[df.Quantity.groupby(price_group).idxmax()]

Answer 2

类似这样的：

from itertools import groupby

x = [[list(n) for m, n in groupby(df['Price'])]][0]
y = [(ind,val) for ind,val in enumerate(x)]
z = [i[0] for i in y for j in i[1]]
df['label'] = z


# it gives you df like this

#    Unnamed: 0  Unnamed: 1 timestamp  Order     Price  Quantity  label
# 0            0  09.10.2019   9:15:42      0  27850.00      2040      0
# 1            1  09.10.2019   9:15:42      0  27850.00      1980      0
# 2            2  09.10.2019   9:15:53      0  27860.85      1800      1
# 3            3  09.10.2019   9:16:54      0  27860.85      2340      1
# 4            4  09.10.2019   9:18:48      0  27860.85      1500      1
# 5            5  09.10.2019   9:21:08      0  27979.00      1840      2
# 6            6  09.10.2019   9:21:08      0  27979.00      2020      2
# 7            7  09.10.2019   9:21:12      0  27850.00      1800      3
# 8            8  09.10.2019   9:21:15      0  27850.00      1580      3
# 9            9  09.10.2019   9:21:21     35  28000.00      1840      4
# 10          10  09.10.2019   9:21:23     34  28000.00      1800      4
# 11          11  09.10.2019   9:28:17      0  28035.00      2020      5
# 12          12  09.10.2019   9:28:18      0  28035.00      1960      5
# 13          13  09.10.2019   9:28:18      0  28035.00      1920      5
# 14          14  09.10.2019   9:28:24      0  28035.00      1940      5
# 15          15  09.10.2019   9:28:24      0  28035.00      1960      5
# 16          16  09.10.2019   9:28:25      0  28000.00      2140      6
# 17          17  09.10.2019   9:28:25      0  28000.00      2020      6
# 18          18  09.10.2019   9:28:26      0  28000.00      2120      6

# then you able to use groupby

df.groupby('label').max()



Out[27]: 
       Unnamed: 0  Unnamed: 1 timestamp  Order     Price  Quantity
label                                                             
0               1  09.10.2019   9:15:42      0  27850.00      2040
1               4  09.10.2019   9:18:48      0  27860.85      2340
2               6  09.10.2019   9:21:08      0  27979.00      2020
3               8  09.10.2019   9:21:15      0  27850.00      1800
4              10  09.10.2019   9:21:23     35  28000.00      1840
5              15  09.10.2019   9:28:24      0  28035.00      2020
6              18  09.10.2019   9:28:26      0  28000.00      2140

Answer 3

这不是最简单的解决方案，但我认为它使正在发生的事情更加明显。我确信它可以被精简为更简洁的代码。

import pandas as pd    

# Generating a similar df
df = pd.DataFrame({'Order'   :[1,2,3,4,5,6,7],
                   'Price'   :[27850.00,27850.00,27860.85,27860.85,27860.85,27979.00,27979.00],
                   'Quantity':[2040,    1980,    1800,    2340    ,1500,    1840,    2020   ] 
                   })

print(df)
print("--------------")

# Get the unique values from the Price column
# This tells us which values we want to select the highest value from 
values = df["Price"].unique()

# Loop through the values, selecting the rows which match each value, one at a time 
for value in values:
    # df["Price"] == value" (Selects all the rows where price equals ONE of the values)
    # For example, the above will give us 3 rows where Price == 27860.85
    # .max() gives us the row with the largest value from Quantity, since the Price column are all equal
    # The above would give us a Series with two values, Price and Quantity. I.e.
    #     Price       27860.85
    #     Quantity     2340.00
    #  ["Quantity"] then selects only the Quantity value and assigns it to highest
    highest = df[df["Price"] == value].max()["Quantity"]
    print(value, "...", highest) 
    # You can, during this loop, build a new dict object to create a new df if desired

或者，更简洁...

# Create a new list in one line
highest = [ df[df["Price"] == value].max()["Quantity"] for value in df["Price"].unique()]
# Add as columns to new df
df1 = pd.DataFrame({
                   'Price'   :df["Price"].unique(),
                   'Quantity':highest
                   })
print(df1) 

使用相同的想法从其他列中为每个唯一的Price 获取适当的值，并将它们添加到新的 df1 中