pandas drop duplicate row value from a specific column - pandas

I want to remove the duplicate row value from a specific column - in this case the column name is "number".
Before:
number qty status
0 10 2 go
1 10 5 nogo
2 4 6 yes
3 3 1 no
4 2 7 go
5 5 2 nah
6 5 6 go
7 5 3 nogo
8 1 10 yes
9 1 10 go
10 5 2 nah
After:
number qty status
0 10 2 go
5 nogo
1 4 6 yes
2 3 1 no
3 2 7 go
4 5 2 nah
6 go
3 nogo
5 1 10 yes
10 go
6 5 2 nah


It is possible replace values to empty string or NaNs by mask with duplicated by new Series a created by comparing column with shifted column with cumsum:
a = df['number'].ne(df['number'].shift()).cumsum()
#for replace ''
df['number'] = df['number'].mask(a.duplicated(), '')
#for replace NaNs
#df['number'] = df['number'].mask(a.duplicated())
print (df)
number qty status
0 10 2 go
1 5 nogo
2 4 6 yes
3 3 1 no
4 2 7 go
5 5 2 nah
6 6 go
7 3 nogo
8 1 10 yes
9 10 go
10 5 2 nah
Detail:
a = df['number'].ne(df['number'].shift()).cumsum()
print (a)
0 1
1 1
2 2
3 3
4 4
5 5
6 5
7 5
8 6
9 6
10 7
Name: number, dtype: int32

Related

pandas dataframe enforce monotically per row

I have a dataframe:
df = 0 1 2 3 4
1 1 3 2 5
4 1 5 7 8
7 1 2 3 9
I want to enforce monotonically per row, to get:
df = 0 1 2 3 4
1 1 3 3 5
4 4 5 7 8
7 7 7 7 9
What is the best way to do so?

Try cummax
out = df.cummax(1)
Out[80]:
0 1 2 3 4
0 1 1 3 3 5
1 4 4 5 7 8
2 7 7 7 7 9

If a column value does not have a certain number of occurances in a dataframe, how to duplicate rows at random until that count is met?

Say that this is what my dataframe looks like
A B
0 1 5
1 4 2
2 3 5
3 3 3
4 3 2
5 2 0
6 4 5
7 2 3
8 4 1
9 5 1
I want every unique value in Column B to occur at least 3 times. So none of the rows with a B value of 5 are duplicated. The row with a column B value of 0 are duplicated twice. And the rest have one of their two rows duplicated at random.
Here is an example desired output
A B
0 1 5
1 4 2
2 3 5
3 3 3
4 3 2
5 2 0
6 4 5
7 2 3
8 4 1
9 5 1
10 4 2
11 2 3
12 2 0
13 2 0
14 4 1
Edit:
The row chosen to be duplicated should be selected at random

To random pick rows, I would use groupby apply with sample on each group. x of lambda is each group of B, so I use reapeat - x.shape[0] to find number of rows need to create. There may be some cases group B already has more rows than 3, so I use np.clip to force negative values to 0. Sample on 0 row is the same as ignore it. Finally, reset_index and append back to df
repeats = 3
df1 = (df.groupby('B').apply(lambda x: x.sample(n=np.clip(repeats-x.shape[0], 0, np.inf)
.astype(int), replace=True))
.reset_index(drop=True))
df_final = df.append(df1).reset_index(drop=True)
Out[43]:
A B
0 1 5
1 4 2
2 3 5
3 3 3
4 3 2
5 2 0
6 4 5
7 2 3
8 4 1
9 5 1
10 2 0
11 2 0
12 5 1
13 4 2
14 2 3

Group counts in new column

I want a new column "group_count". This shows me in how many groups in total the attribute occurs.
Group Attribute group_count
0 1 10 4
1 1 10 4
2 1 10 4
3 2 10 4
4 2 20 1
5 3 30 1
6 3 10 4
7 4 10 4
I tried to groupby Group and attributes and then transform by using count
df["group_count"] = df.groupby(["Group", "Attributes"])["Attributes"].transform("count")
Group Attribute group_count
0 1 10 3
1 1 10 3
2 1 10 3
3 2 10 1
4 2 20 1
5 3 30 1
6 3 10 1
7 4 10 1
But it doesnt work

Use df.drop_duplicates(['Group','Attribute']) to get unique Attribute per group , then groupby on Atttribute to get count of Group, finally map with original Attribute column.
m=df.drop_duplicates(['Group','Attribute'])
df['group_count']=df['Attribute'].map(m.groupby('Attribute')['Group'].count())
print(df)
Group Attribute group_count
0 1 10 4
1 1 10 4
2 1 10 4
3 2 10 4
4 2 20 1
5 3 30 1
6 3 10 4
7 4 10 4

Use DataFrameGroupBy.nunique with transform:
df['group_count1'] = df.groupby('Attribute')['Group'].transform('nunique')
print (df)
Group Attribute group_count group_count1
0 1 10 4 4
1 1 10 4 4
2 1 10 4 4
3 2 10 4 4
4 2 20 1 1
5 3 30 1 1
6 3 10 4 4
7 4 10 4 4

Comparing two dataframe and output the index of the duplicated row once

I need help with comparing two dataframes. For example:
The first dataframe is
df_1 =
0 1 2 3 4 5
0 1 1 1 1 1 1
1 2 2 2 2 2 2
2 3 3 3 3 3 3
3 4 4 4 4 4 4
4 2 2 2 2 2 2
5 5 5 5 5 5 5
6 1 1 1 1 1 1
7 6 6 6 6 6 6
The second dataframe is
df_2 =
0 1 2 3 4 5
0 1 1 1 1 1 1
1 2 2 2 2 2 2
2 3 3 3 3 3 3
3 4 4 4 4 4 4
4 5 5 5 5 5 5
5 6 6 6 6 6 6
May I know if there is a way (without using for loop) to find the index of the rows of df_1 that have the same row values of df_2. In the example above, my expected output is below
index =
0
1
2
3
5
7
The size of the column of the "index" variable above should have the same column size of df_2.
If the same row of df_2 repeated in df_1 more than once, I only need the index of the first appearance, thats why I don't need the index 4 and 6.
Please help. Thank you so much!
Tommy

Use DataFrame.merge with DataFrame.drop_duplicates and DataFrame.reset_index for convert index to column for avoid lost index values, last select column called index:
s = df_2.merge(df_1.drop_duplicates().reset_index())['index']
print (s)
0 0
1 1
2 2
3 3
4 5
5 7
Name: index, dtype: int64
Detail:
print (df_2.merge(df_1.drop_duplicates().reset_index()))
0 1 2 3 4 5 index
0 1 1 1 1 1 1 0
1 2 2 2 2 2 2 1
2 3 3 3 3 3 3 2
3 4 4 4 4 4 4 3
4 5 5 5 5 5 5 5
5 6 6 6 6 6 6 7

Check the solution
df1=pd.DataFrame({'0':[1,2,3,4,2,5,1,6],
'1':[1,2,3,4,2,5,1,6],
'2':[1,2,3,4,2,5,1,6],
'3':[1,2,3,4,2,5,1,6],
'4':[1,2,3,4,2,5,1,6],
'5':[1,2,3,4,2,5,1,6]})
df1=pd.DataFrame({'0':[1,2,3,4,5,6],
'1':[1,2,3,4,5,66],
'2':[1,2,3,4,5,6],
'3':[1,2,3,4,5,66],
'4':[1,2,3,4,5,6],
'5':[1,2,3,4,5,6]})
df1[df1.isin(df2)].index.values.tolist()
### Output
[0, 1, 2, 3, 4, 5, 6, 7]

pandas: bin data into specific number of bins of specific size

I would like to bin a dataframe by the values in a single column into bins of a specific size and number.
Here is an example df:
df= pd.DataFrame(np.random.randint(0,10000,size=(10000, 4)), columns=list('ABCD'))
Say I want to bin by column D, I will first sort the data:
df.sort('D')
I would now wish to bin so that the first if bin size is 50 and bin number is 100, the first 50 values will go into bin 1, the next into bin 2, and so on and so forth. Any remaining values after the twenty bins should all go into the final bin. Is there anyway of doing this?
EDIT:
Here is a sample input:
x = pd.DataFrame(np.random.randint(0,10,size=(10, 4)), columns=list('ABCD'))
And here is the expected output:
A B C D bin
0 6 8 6 5 3
1 5 4 9 1 1
2 5 1 7 4 3
3 6 3 3 3 2
4 2 5 9 3 2
5 2 5 1 3 2
6 0 1 1 0 1
7 3 9 5 8 3
8 2 4 0 1 1
9 6 4 5 6 3
As an extra aside, is it also possible to bin any equal values in the same bin? So for example, say I have bin 1 which contains values, 0,1,1 and then bin 2 contains 1,1,2. Is there any way of putting those two 1 values in bin 2 into bin 1? This will create very uneven bin sizes but this is not an issue.

It seems you need floor divide np.arange and then assign to new column:
idx = df['D'].sort_values().index
df['b'] = pd.Series(np.arange(len(df)) // 3 + 1, index = idx)
print (df)
A B C D bin b
0 6 8 6 5 3 3
1 5 4 9 1 1 1
2 5 1 7 4 3 3
3 6 3 3 3 2 2
4 2 5 9 3 2 2
5 2 5 1 3 2 2
6 0 1 1 0 1 1
7 3 9 5 8 3 4
8 2 4 0 1 1 1
9 6 4 5 6 3 3
Detail:
print (np.arange(len(df)) // 3 + 1)
[1 1 1 2 2 2 3 3 3 4]
EDIT:
I create another question about problem with last values here:
N = 3
idx = df['D'].sort_values().index
#one possible solution, thanks divakar
def replace_irregular_groupings(a, N):
n = len(a)
m = N*(n//N)
if m!=n:
a[m:] = a[m-1]
return a
idx = df['D'].sort_values().index
arr = replace_irregular_groupings(np.arange(len(df)) // N + 1, N)
df['b'] = pd.Series(arr, index = idx)
print (df)
A B C D bin b
0 6 8 6 5 3 3
1 5 4 9 1 1 1
2 5 1 7 4 3 3
3 6 3 3 3 2 2
4 2 5 9 3 2 2
5 2 5 1 3 2 2
6 0 1 1 0 1 1
7 3 9 5 8 3 3
8 2 4 0 1 1 1
9 6 4 5 6 3 3