在pandas python中从具有复杂标准的数据框中过滤数据

Question

这是我的数据框

import pandas as pd

data=pd.DataFrame({'vehicle':['car','car','car','car','car','car','bus','bus','bus','bus','bus','bus','car','car','car','car','car','car','bus','bus','bus','bus','bus','bus'],
'expecteddate':['2/24/2022','2/24/2022','3/15/2022','3/15/2022','4/20/2022','4/20/2022','2/24/2022','2/24/2022','3/15/2022','3/15/2022','4/20/2022','4/20/2022','2/24/2022','2/24/2022','3/15/2022','3/15/2022','4/20/2022','4/20/2022','2/24/2022','2/24/2022','3/15/2022','3/15/2022','4/20/2022','4/20/2022'],'range':[240,240,240,240,240,240,300,300,300,300,300,300,240,240,240,240,240,240,300,300,300,300,300,300],'color':['blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red','blue','red'],'discount':[70,80,90,60,40,50,120,110,130,140,80,90,60,40,50,30,70,45,130,100,140,120,90,30],'date':['2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/18/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022','2/17/2022']})
print(data)

数据框中的数据：

   vehicle  expecteddate  range color  discount       date
0      car    2/24/2022    240  blue        70  2/18/2022
1      car    2/24/2022    240   red        80  2/18/2022
2      car    3/15/2022    240  blue        90  2/18/2022
3      car    3/15/2022    240   red        60  2/18/2022
4      car    4/20/2022    240  blue        40  2/18/2022
5      car    4/20/2022    240   red        50  2/18/2022
6      bus    2/24/2022    300  blue       120  2/18/2022
7      bus    2/24/2022    300   red       110  2/18/2022
8      bus    3/15/2022    300  blue       130  2/18/2022
9      bus    3/15/2022    300   red       140  2/18/2022
10     bus    4/20/2022    300  blue        80  2/18/2022
11     bus    4/20/2022    300   red        90  2/18/2022
12     car    2/24/2022    240  blue        60  2/17/2022
13     car    2/24/2022    240   red        40  2/17/2022
14     car    3/15/2022    240  blue        50  2/17/2022
15     car    3/15/2022    240   red        30  2/17/2022
16     car    4/20/2022    240  blue        70  2/17/2022
17     car    4/20/2022    240   red        45  2/17/2022
18     bus    2/24/2022    300  blue       130  2/17/2022
19     bus    2/24/2022    300   red       100  2/17/2022
20     bus    3/15/2022    300  blue       140  2/17/2022
21     bus    3/15/2022    300   red       120  2/17/2022
22     bus    4/20/2022    300  blue        90  2/17/2022
23     bus    4/20/2022    300   red        30  2/17/2022

从这个数据框中，我们有两辆车，三个预期日期，范围，两种颜色，折扣和日期。 我们必须找到折扣中的最小值和我们在折扣列中获得最小值的日期，分成两个新列，即 mindisc 列和 mindate，应该根据最新日期保存在新列中，这应该根据车辆过滤,expecteddate,range,color and date

我们必须在两个日期的折扣列中找到最小值（所有日期，因为我们有很多日期，不限于两个日期）2/18/2022,2/17/2022 基于相同的颜色、范围、预期日期和车辆

最后这个最小值添加到最晚日期的 mindisc 列和 min date 出现在最晚日期行的 minate 列的相应日期

输出应该是这样的

 country expecteddate  range color  discount       date  mindisc    mindate
0      car    2/24/2022    240  blue        70  2/18/2022   60       2/17/2022
1      car    2/24/2022    240   red        80  2/18/2022   40       2/17/2022
2      car    3/15/2022    240  blue        90  2/18/2022   50       2/17/2022
3      car    3/15/2022    240   red        60  2/18/2022   30       2/17/2022
4      car    4/20/2022    240  blue        40  2/18/2022   40       2/18/2022
5      car    4/20/2022    240   red        50  2/18/2022   45       2/17/2022
6      bus    2/24/2022    300  blue       120  2/18/2022   120      2/18/2022
7      bus    2/24/2022    300   red       110  2/18/2022   100      2/17/2022
8      bus    3/15/2022    300  blue       130  2/18/2022   130      2/18/2022
9      bus    3/15/2022    300   red       140  2/18/2022   120      2/17/2022
10     bus    4/20/2022    300  blue        80  2/18/2022   80       2/18/2022
11     bus    4/20/2022    300   red        90  2/18/2022   30       2/17/2022
12     car    2/24/2022    240  blue        60  2/17/2022
13     car    2/24/2022    240   red        40  2/17/2022
14     car    3/15/2022    240  blue        50  2/17/2022
15     car    3/15/2022    240   red        30  2/17/2022
16     car    4/20/2022    240  blue        70  2/17/2022
17     car    4/20/2022    240   red        45  2/17/2022
18     bus    2/24/2022    300  blue       130  2/17/2022
19     bus    2/24/2022    300   red       100  2/17/2022
20     bus    3/15/2022    300  blue       140  2/17/2022
21     bus    3/15/2022    300   red       120  2/17/2022
22     bus    4/20/2022    300  blue        90  2/17/2022
23     bus    4/20/2022    300   red        30  2/17/2022

车辆不限于汽车和公共汽车等两个，它有很多车辆，数据在车辆和范围列中的行并不总是相等，日期不限于两个日期

Answer 1

这是我的方法。让我们首先将"date" 列视为一个日期时间对象：

df["date"] = pd.to_datetime(df["date"])

现在，我们可以按照您的描述对数据进行分组，以找到达到最低折扣的行：

common_groupby = df.groupby(["color", "range", "expecteddate", "vehicle"])

现在，让我们找出每个组出现最小折扣和最大日期的行：

source_idx = common_groupby["discount"].idxmin()
target_idx = common_groupby["date"].idxmax()
# Use df.loc[idx] to see the rows where the minimum discount is reached

现在，我们可以使用.loc 将最低折扣值和相应日期（来自source_idx 行）添加到正确的行（来自target_src）：

df.loc[target_idx, "mindisc"] = df.loc[source_idx, "discount"].values
df.loc[target_idx, "mindate"] = df.loc[source_idx, "date"].values

如您所见，我们仅更改达到最低折扣的行中的值 (idx)。这是这些操作的输出：

   vehicle expecteddate  range color  discount       date  mindisc    mindate
0      car    2/24/2022    240  blue        70 2022-02-18     60.0 2022-02-17
1      car    2/24/2022    240   red        80 2022-02-18     40.0 2022-02-17
2      car    3/15/2022    240  blue        90 2022-02-18     50.0 2022-02-17
3      car    3/15/2022    240   red        60 2022-02-18     30.0 2022-02-17
4      car    4/20/2022    240  blue        40 2022-02-18     40.0 2022-02-18
5      car    4/20/2022    240   red        50 2022-02-18     45.0 2022-02-17
6      bus    2/24/2022    300  blue       120 2022-02-18    120.0 2022-02-18
7      bus    2/24/2022    300   red       110 2022-02-18    100.0 2022-02-17
8      bus    3/15/2022    300  blue       130 2022-02-18    130.0 2022-02-18
9      bus    3/15/2022    300   red       140 2022-02-18    120.0 2022-02-17
10     bus    4/20/2022    300  blue        80 2022-02-18     80.0 2022-02-18
11     bus    4/20/2022    300   red        90 2022-02-18     30.0 2022-02-17
12     car    2/24/2022    240  blue        60 2022-02-17      NaN        NaT
13     car    2/24/2022    240   red        40 2022-02-17      NaN        NaT
14     car    3/15/2022    240  blue        50 2022-02-17      NaN        NaT
15     car    3/15/2022    240   red        30 2022-02-17      NaN        NaT
16     car    4/20/2022    240  blue        70 2022-02-17      NaN        NaT
17     car    4/20/2022    240   red        45 2022-02-17      NaN        NaT
18     bus    2/24/2022    300  blue       130 2022-02-17      NaN        NaT
19     bus    2/24/2022    300   red       100 2022-02-17      NaN        NaT
20     bus    3/15/2022    300  blue       140 2022-02-17      NaN        NaT
21     bus    3/15/2022    300   red       120 2022-02-17      NaN        NaT
22     bus    4/20/2022    300  blue        90 2022-02-17      NaN        NaT
23     bus    4/20/2022    300   red        30 2022-02-17      NaN        NaT

---

另一种可能的解决方案试图通过排序然后删除重复两次来提高性能（受此问题启发：Select the max row per group - pandas performance issue）：

# Min discount rows
source_df = df.sort_values(by=["discount"], ascending=True, kind='mergesort').drop_duplicates(["color", "range", "expecteddate", "vehicle"])
source_df = source_df.rename(columns={"date": "mindate", "discount": "mindisc"})

# Max date rows
target_df = df.reset_index().sort_values(by=["date"], ascending=False, kind="mergesort").drop_duplicates(["color", "range", "expecteddate", "vehicle"])

# Put min discount values into max date rows
df.loc[target_df["index"], ["mindisc", "mindate"]] = source_df[["mindisc", "mindate"]].values

Answer 2

我没有很好地解决问题，但您可以尝试以下方法：

new_df = df[(df['date'] == '2/18/2022') & (df['color'] == 'blue') & (df['vehicle'] == 'car')]

然后：

new_df['discount'].min()