Question

我有以下df

col_1   col_2
1   1
1   2
1   3
1   6
1   8
1   11
1   12
1   19
1   24
1   1
1   1
1   2
1   2
1   3 
1   3
2   1
2   2
2   4
2   6
2   7
2   11
2   13
2   16
2   19
2   1
2   2
2   3

我想对col_1进行分组，并替换col_2中在 19之后出现的值1、2、3，并将其更改为25， 26，27。

预期输出：

col_1   col_2
1   1
1   2
1   3
1   6
1   8
1   11
1   12
1   19
1   24
1   25
1   25
1   26
1   26
1   27
1   27
2   1
2   2
2   4
2   6
2   7
2   11
2   13
2   16
2   19
2   25
2   26
2   27

我想知道如何使用熊猫来做到这一点。

谢谢

编辑1：

我真正的df

ContextID   BacksGas_Flow_sccm  StepID
7289973 1.953125    1
7289973 2.05078125  2
7289973 2.05078125  2
7289973 2.05078125  2
7289973 1.953125    2
7289973 1.7578125   2
7289973 1.7578125   2
7289973 1.85546875  2
7289973 1.7578125   2
7289973 9.08203125  5
7289973 46.19140625 5
7289973 46.19140625 5
7289973 46.19140625 5
7289973 46.19140625 5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  5
7289973 46.6796875  7
7289973 46.6796875  7
7289973 46.6796875  7
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  12
7289973 46.6796875  15
7289973 46.6796875  15
7289973 46.6796875  16
7289973 46.6796875  16
7289973 46.6796875  17
7289973 25.09765625 19
7289973 45.99609375 19
7289973 59.08203125 19
7289973 61.81640625 19
7289973 62.59765625 19
7289973 63.671875   19
7289973 65.625  19
7289973 66.69921875 19
7289973 67.3828125  19
7289973 67.3828125  19
7289973 67.67578125 19
7289973 68.26171875 19
7289973 69.04296875 19
7289973 69.82421875 19
7289973 69.82421875 19
7289973 70.8984375  19
7289973 70.8984375  19
7289973 70.8984375  19
7289973 70.8984375  19
7289973 71.6796875  19
7289973 71.6796875  19
7289973 72.55859375 19
7289973 72.55859375 19
7289973 72.55859375 19
7289973 72.55859375 19
7289973 72.55859375 19
7289973 72.55859375 19
7289973 73.33984375 19
7289973 73.33984375 19
7289973 73.33984375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 73.2421875  19
7289973 73.2421875  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.0234375  19
7289973 74.90234375 19
7289973 74.90234375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 75          19
7289973 75          19
7289973 75          19
7289973 74.21875    19
7289973 74.21875    19
7289973 74.21875    19
7289973 75          19
7289973 75          19
7289973 75          19
7289973 75          19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.12109375 19
7289973 74.90234375 19
7289973 6.4453125   24
7289973 3.515625    24
7289973 2.5390625   24
7289973 2.05078125  24
7289973 2.05078125  24
7289973 2.05078125  24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.953125    24
7289973 2.05078125  24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 2.05078125  24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.7578125   24
7289973 1.66015625  24
7289973 1.7578125   24
7289973 1.7578125   24
7289973 1.7578125   24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.953125    24
7289973 1.7578125   24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.7578125   24
7289973 1.85546875  24
7289973 1.85546875  24
7289973 1.7578125   24
7289973 1.7578125   1
7289973 1.85546875  1
7289973 1.85546875  1
7289973 1.85546875  2
7289973 1.7578125   2
7289973 1.953125    2
7289973 1.953125    2
7289973 1.85546875  2
7289973 1.85546875  3
7289973 1.85546875  3
7289973 1.85546875  3
7289973 1.953125    3
7289973 1.85546875  3
7289973 1.953125    3
7289973 1.85546875  3
7289973 1.7578125   3
7289973 1.85546875  3
7289973 1.85546875  3
7289973 1.7578125   3
7289973 1.85546875  3

Answer 1

一种方法是创建一个字典来replace col_2中的值。为了仅替换出现在19，GroupBy之后的内容，请检查相等性并使用cumsum对数据帧执行布尔索引：

map_ = {1:25, 2:26, 3:27}
cs = df.col_2.eq(19).groupby(df.col_1).cumsum()
update = df.loc[cs].col_2.replace(map_)
df.loc[update.index, 'col_2'] = update

 col_1  col_2
0       1      1
1       1      2
2       1      3
3       1      6
4       1      8
5       1     11
6       1     12
7       1     19
8       1     25
9       1     26
10      1     27
11      2      1
12      2      2
13      2      4
14      2      6
15      2      7
16      2     11
17      2     13
18      2     16
19      2     19
20      2     25
21      2     26
22      2     27

Answer 2

我的尝试：

def fill19(x):
    # x.shift()==19 marks all 1's after 19's
    # rolling(3) marks three numbers after 19's
    filters = (x.shift()==19).rolling(3).sum().fillna(0).astype(bool)
    x[filters] += 24
    return x

df.col2 = df.groupby('col_1').col_2.apply(fill19)

0      1
1      2
2      3
3      6
4      8
5     11
6     12
7     19
8     25
9     26
10    27
11     1
12     2
13     4
14     6
15     7
16    11
17    13
18    16
19    19
20    25
21    26
22    27
Name: col_2, dtype: int64

Answer 3

尝试以下for循环，它可以完成您所需要的操作：

for i in df['col_1'].unique():
     ix = np.argwhere((df['col_1'] == i) & (df['col_2'] == 19 ))
     df.loc[ix[0][0]+1, 'col_2'] = 25
     df.loc[ix[0][0]+2, 'col_2'] = 26
     df.loc[ix[0][0]+3, 'col_2'] = 27

Answer 4

守旧派循环赛

map_ = {1: 25, 2: 26, 3: 27}

d = {}  # Tracks if 19 has been seen yet

for i, c1, c2 in df.itertuples():

    if d.setdefault(c1, False):
        df.at[i, 'col_2'] = map_.get(c2, c2)

    d[c1] |= c2 == 19

使用`np.logigcal_or`

m = df.col_2.eq(19)
m = m.groupby(df.col_1).transform(np.logical_or.accumulate) ^ m
df.assign(col_2=df.col_2 + m * 24)

    col_1  col_2
0       1      1
1       1      2
2       1      3
3       1      6
4       1      8
5       1     11
6       1     12
7       1     19
8       1     25
9       1     26
10      1     27
11      2      1
12      2      2
13      2      4
14      2      6
15      2      7
16      2     11
17      2     13
18      2     16
19      2     19
20      2     25
21      2     26
22      2     27

Answer 5

基于真实的DataFrame，您可以执行以下操作：

df['StepID'] = df.groupby('ContextID')['StepID'].apply(
    lambda x: x + (x < x.shift(1)).cumsum()*24)

print(df.tail(25))

输出：

     ContextID  BacksGas_Flow_sccm  StepID
138    7289973            1.855469      24
139    7289973            1.757812      24
140    7289973            1.855469      24
141    7289973            1.855469      24
142    7289973            1.757812      24
143    7289973            1.757812      25
144    7289973            1.855469      25
145    7289973            1.855469      25
146    7289973            1.855469      26
147    7289973            1.757812      26
148    7289973            1.953125      26
149    7289973            1.953125      26
150    7289973            1.855469      26
151    7289973            1.855469      27
152    7289973            1.855469      27
153    7289973            1.855469      27
154    7289973            1.953125      27
155    7289973            1.855469      27
156    7289973            1.953125      27
157    7289973            1.855469      27
158    7289973            1.757812      27
159    7289973            1.855469      27
160    7289973            1.855469      27
161    7289973            1.757812      27
162    7289973            1.855469      27

P.S。在您的示例数据中，只有一个ContextID，但是我假设整个数据集中也可能还有其他groupby

更新：如果您仅需要将每个24的{{1}}之后的值每24递增一次（我将新值保存到{{1} }列以显示转换前后）：

ContextID

输出：

StepID_new

如何根据特定条件替换列中的某些值？

5 个答案:

守旧派循环赛

使用`np.logigcal_or`

如何根据特定条件替换列中的某些值？

5 个答案:

守旧派循环赛

使用np.logigcal_or

使用`np.logigcal_or`