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How to mark 'duplicated sequence' in pandas?


How to mark 'duplicated sequence' in pandas?

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How to mark 'duplicated sequence' in pandas?
Tag : python , By : Kristian Hofslaeter
Date : November 25 2020, 03:01 PM

may help you . You need shift with cumsum for helper Series where apply duplicated:
s = pd.Series([1,2,1,1,2,3,3,2,4,2,2,1])
print (s.ne(s.shift()).cumsum().duplicated(keep=False).values)
[False False  True  True False  True  True False False  True  True False]
print (s.ne(s.shift()).cumsum())
0     1
1     2
2     3
3     3
4     4
5     5
6     5
7     6
8     7
9     8
10    8
11    9
dtype: int32

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pandas: aggregate a column to create a non-duplicated sequence


Tag : python-3.x , By : ganok_tor
Date : March 29 2020, 07:55 AM
should help you out If you allow non consecutive duplicates, you must filter carefully. A way to do that :
def filter(l):
    l.append(None)
    return ','.join([x for (i,x) in enumerate (l[:-1])
    if l[i] != l[i+1]])

out=df.groupby('name')['color'].apply(list).apply(filter)
name
Ann     green,orange,red,black
Dan          blue,green,yellow
John           blue,yellow,red
Name: color, dtype: object

Python/pandas: sequence to series (transform a sequence so that each element of the series is a sum of sequence elements


Tag : python , By : barefootChild
Date : March 29 2020, 07:55 AM
I hope this helps . Since you haven't provided the expected DF, I'm guessing you want to compute successive difference between cells, take it's absolute value and then perform the rolling summation whose window length would vary depending on the lookback value.
res = items.diff().abs().rolling(window=lookback).sum()

Pandas merge with duplicated key - removing duplicated rows or preventing it's creation


Tag : python , By : Vasiliy
Date : March 29 2020, 07:55 AM
wish help you to fix your issue I suggest create new helper column for count id values by cumcount and then merge by this values:
df1['g'] = df1.groupby('id').cumcount()
df2['g'] = df2.groupby('id').cumcount()

merged_table = pd.merge(df1,df2,on=["id", 'g'],how='outer')
print (merged_table)
    Name  amount_x    id  g Category  amount_y
0   John    500.25  GH10  0     Food    500.25
1  Helen   1250.00  GH11  0   Travel   1250.00
2   Adam    432.54  GH11  1     Food    432.54
3  Sarah    567.12  GH12  0      NaN       NaN
merged_table = pd.merge(df1,df2,on=["id", 'g'],how='outer').drop('g', axis=1)
print (merged_table)
    Name  amount_x    id Category  amount_y
0   John    500.25  GH10     Food    500.25
1  Helen   1250.00  GH11   Travel   1250.00
2   Adam    432.54  GH11     Food    432.54
3  Sarah    567.12  GH12      NaN       NaN 
print (df1)
    Name   amount    id  g
0   John   500.25  GH10  0
1  Helen  1250.00  GH11  0
2   Adam   432.54  GH11  1
3  Sarah   567.12  GH12  0

print (df2)
  Category   amount    id  g
0     Food   500.25  GH10  0
1   Travel  1250.00  GH11  0
2     Food   432.54  GH11  1

How to count each x entries and mark the occurence of this sequence with a value in a pandas dataframe?


Tag : pandas , By : Chris Hubbard
Date : March 29 2020, 07:55 AM
This might help you General solution is create numpy array by np.arange, then use integer division by 4 and add 1, because python count from 0:
df['C'] = np.arange(len(df)) // 4 + 1
print (df)
     A    B  C
0    1  100  1
1    2  102  1
2    3  103  1
3    4  104  1
4    5  105  2
5    6  106  2
6    7  108  2
7    8  109  2
8    9  110  3
9   10  112  3
10  11  113  3
11  12  115  3
12  13  116  4
13  14  118  4
14  15  120  4
15  16  121  4

How to count and mark the occurence of a sequence of a value in a pandas dataframe?


Tag : pandas , By : helloedwin
Date : March 29 2020, 07:55 AM
wish of those help I want to create a column C (based on B) which counts each beginning of a series from '100' in B. I have the following pandas data frame: , Use:
df['C'] = (df['B'].shift(-1).eq(100) & df['B'].ne(100)).cumsum() * df['B'].eq(100)
print (df)
     A    B  C
0    1    0  0
1    2    0  0
2    3  100  1
3    4  100  1
4    5  100  1
5    6    0  0
6    7    0  0
7    8  100  2
8    9  100  2
9   10  100  2
10  11  100  2
11  12    0  0
12  13    0  0
13  14    0  0
14  15  100  3
15  16  100  3
df = df.assign(shifted = df['B'].shift(-1).eq(100),
               chained = df['B'].shift(-1).eq(100) & df['B'].ne(100),
               cumsum = (df['B'].shift(-1).eq(100) & df['B'].ne(100)).cumsum(),
               eq_100 = df['B'].eq(100),
               C = (df['B'].shift(-1).eq(100) & df['B'].ne(100)).cumsum() * df['B'].eq(100))
print (df)
     A    B  shifted  chained  cumsum  eq_100  C
0    1    0    False    False       0   False  0
1    2    0     True     True       1   False  0
2    3  100     True    False       1    True  1
3    4  100     True    False       1    True  1
4    5  100    False    False       1    True  1
5    6    0    False    False       1   False  0
6    7    0     True     True       2   False  0
7    8  100     True    False       2    True  2
8    9  100     True    False       2    True  2
9   10  100     True    False       2    True  2
10  11  100    False    False       2    True  2
11  12    0    False    False       2   False  0
12  13    0    False    False       2   False  0
13  14    0     True     True       3   False  0
14  15  100     True    False       3    True  3
15  16  100    False    False       3    True  3
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