Identify consecutive same values in Pandas Dataframe, with a Groupby

You can try this; 1) Create an extra group variable with df.value.diff().ne(0).cumsum() to denote the value changes; 2) use transform('size') to calculate the group size and compare with three, then you get the flag column you need:

df['flag'] = df.value.groupby([df.id, df.value.diff().ne(0).cumsum()]).transform('size').ge(3).astype(int) 
df

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Break downs:

1) diff is not equal to zero (which is literally what df.value.diff().ne(0) means) gives a condition True whenever there is a value change:

df.value.diff().ne(0)
#0      True
#1     False
#2      True
#3      True
#4     False
#5     False
#6      True
#7     False
#8     False
#9     False
#10     True
#11     True
#12     True
#13    False
#14    False
#15     True
#16    False
#17     True
#18    False
#19    False
#20    False
#21    False
#Name: value, dtype: bool

2) Then cumsum gives a non descending sequence of ids where each id denotes a consecutive chunk with same values, note when summing boolean values, True is considered as one while False is considered as zero:

df.value.diff().ne(0).cumsum()
#0     1
#1     1
#2     2
#3     3
#4     3
#5     3
#6     4
#7     4
#8     4
#9     4
#10    5
#11    6
#12    7
#13    7
#14    7
#15    8
#16    8
#17    9
#18    9
#19    9
#20    9
#21    9
#Name: value, dtype: int64

3) combined with id column, you can group the data frame, calculate the group size and get the flag column.

See EDIT2 for a more robust solution

Same result, but a little bit faster:

labels = (df.value != df.value.shift()).cumsum()
df['flag'] = (labels.map(labels.value_counts()) >= 3).astype(int)

    id  value  flag
0    1      2     0
1    1      2     0
2    1      3     0
3    1      2     1
4    1      2     1
5    1      2     1
6    1      3     1
7    1      3     1
8    1      3     1
9    1      3     1
10   2      1     0
11   2      4     0
12   2      1     1
13   2      1     1
14   2      1     1
15   2      4     0
16   2      4     0
17   2      1     1
18   2      1     1
19   2      1     1
20   2      1     1
21   2      1     1

Where:

1. df.value != df.value.shift() gives the value change
2. cumsum() creates "labels" for each group of same value
3. labels.value_counts() counts the occurrences of each label
4. labels.map(...) replaces labels by the counts computed above
5. >= 3 creates a boolean mask on count value
6. astype(int) casts the booleans to int

In my hands it give 1.03ms on your df, compared to 2.1ms for Psidoms' approach. But mine is not one-liner.

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EDIT:

A mix between both approaches is even faster

labels = df.value.diff().ne(0).cumsum()
df['flag'] = (labels.map(labels.value_counts()) >= 3).astype(int)

Gives 911µs with your sample df.

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EDIT2: correct solution to account for id change, as pointed by @clg4

labels = (df.value.diff().ne(0) | df.id.diff().ne(0)).cumsum()
df['flag'] = (labels.map(labels.value_counts()) >= 3).astype(int)

Where ... | df.id.diff().ne(0) increment the label where the id changes

This works even with same value on id change (tested with value 3 on index 10) and takes 1.28ms

EDIT3: Better explanations

Take the case where index 10 has value 3. df.id.diff().ne(0)

data={'id':[1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2],
      'value':[2,2,3,2,2,2,3,3,3,3,3,4,1,1,1,4,4,1,1,1,1,1]}
df=pd.DataFrame.from_dict(data)

df['id_diff'] = df.id.diff().ne(0).astype(int)
df['val_diff'] = df.value.diff().ne(0).astype(int)
df['diff_or'] = (df.id.diff().ne(0) | df.value.diff().ne(0)).astype(int)
df['labels'] = df['diff_or'].cumsum()

     id  value  id_diff  val_diff  diff_or  labels
 0    1      2        1         1        1       1
 1    1      2        0         0        0       1
 2    1      3        0         1        1       2
 3    1      2        0         1        1       3
 4    1      2        0         0        0       3
 5    1      2        0         0        0       3
 6    1      3        0         1        1       4
 7    1      3        0         0        0       4
 8    1      3        0         0        0       4
 9    1      3        0         0        0       4
>10   2      3        1    |    0    =   1       5 <== label increment
 11   2      4        0         1        1       6
 12   2      1        0         1        1       7
 13   2      1        0         0        0       7
 14   2      1        0         0        0       7
 15   2      4        0         1        1       8
 16   2      4        0         0        0       8
 17   2      1        0         1        1       9
 18   2      1        0         0        0       9
 19   2      1        0         0        0       9
 20   2      1        0         0        0       9
 21   2      1        0         0        0       9

The | is operator "bitwise-or", which gives True as long as one of the elements is True. So if there is no diff in value where the id changes, the | reflects the id change. Otherwise it changes nothing. When .cumsum() is performed, the label is incremented where the id changes, so the value 3 at index 10 is not grouped with values 3 from indexes 6-9.