比较 pandas 中 groupby 中的列值

Question

我的 dataframe 看起来像这样

Time    Name    price   Profit   
5:25    A        150       15
5:25    B        250       10
5:25    C        200       20
5:30    A        200       25
5:30    B        150       20
5:30    C        210       25
5:35    A        180       15
5:35    B        200       30
5:35    C        200       10 
5:40    A        150       20 
5:40    B        260       15 
5:40    C        220       10   

我想要 output 应该是这样的：

Time    Name    price  profit    diff_price   diff_profit      
5:25    A        150     15         0            0
5:25    B        250     10         0            0
5:25    C        200     20         0            0
5:30    A        200     25        50            10
5:30    B        150     20        -100          10
5:30    C        210     25         10            5
5:35    A        180     15         20            -10
5:35    B        200     30         50            10
5:35    C        200     10         -10           -15
5:40    A        150     20         -30           5
5:40    B        260     35          60           5
5:40    C        220     15          20           5

我需要比较以前的 groupby 值是否大于以前的值，比如

A、B 和 C 的差是否大于以前的值。 如果条件匹配，它必须显示名称：

从上面的时间 5:40 可以看出，B 的 diff_price 和 diff_profit 大于所有之前的时间列值

所以 output 应该打印成：B

我的代码看起来像

df.groupby(['Time','Price'])
df['diff_price']=df.groupby(['Time','Price']).price.diff().fillna(0)
df['diff_profit']=df.groupby(['Time','Price']).profit.diff().fillna(0)

那么如何在值之间进行比较以获得所需的 output 来显示是：B

Answer 1

IIUC，根据Name列计算diff_price和diff_profit然后根据您的条件修补最后一组时间：

df[['diff_price', 'diff_profit']] = df.groupby('Name')[['price', 'profit']] \
                                      .diff().fillna(0)

mask = df['Time'].eq(df['Time'].max())
df.loc[mask, 'diff_profit'] = df.loc[mask, 'diff_profit'].max()

输出：

>>> df
    Time Name  price  profit  diff_price  diff_profit
0   5:25    A    150      15         0.0          0.0
1   5:25    B    250      10         0.0          0.0
2   5:25    C    200      20         0.0          0.0
3   5:30    A    200      25        50.0         10.0
4   5:30    B    150      20      -100.0         10.0
5   5:30    C    210      25        10.0          5.0
6   5:35    A    180      15       -20.0        -10.0
7   5:35    B    200      30        50.0         10.0
8   5:35    C    200      10       -10.0        -15.0
9   5:40    A    150      20       -30.0          5.0
10  5:40    B    260      15        60.0          5.0
11  5:40    C    220      10        20.0          5.0

Answer 2

您可以同时解决这个问题的一组（“名称”）：

# Let's iterate the dataframe by grouping by "Name"
for name, group_df in df.groupby(["Name"]):
    # Make sure that the rows are sorted by time
    group_df = group_df.sort_values("Time")
    # Calculate difference between each row (diff = bottom - top)
    group_df[["diff_price", "diff_profit"]] = group_df[["price", "Profit"]].shift(1) - group_df[["price", "Profit"]]
    # Fill the first value with 0 instead of NaN (as in your sample input)
    group_df = group_df.fillna(0)

    # Let's see if the maximum diff_price is reached at the end
    *previous_values, last_value = group_df["diff_price"]
    if last_value >= max(previous_values):
        print(f"Max price diff reached at '{name}'")
        print(group_df.tail(1))

    # Again, but let's checkout the diff_profit
    *previous_values, last_value = group_df["diff_profit"]
    if last_value >= max(previous_values):
        print(f"Max profit diff reached at '{name}'")
        print(group_df.tail(1))

这是我为您的示例输入获得的 output：

Max price diff reached: A
   Time Name  price  Profit  diff_price  diff_profit
9  5:40    A    150      20        30.0         -5.0
Max profit diff reached: B
    Time Name  price  Profit  diff_price  diff_profit
10  5:40    B    260      15       -60.0         15.0