I have a data frame that looks like the following
df = pd.DataFrame({'group':[1,1,2,2,2],'time':[1,2,3,4,5],'C':[6,7,8,9,10]})
group time C
0 1 1 6
1 1 2 7
2 2 3 8
3 2 4 9
4 2 5 10
and I'm looking to label the first element (in terms of time) in each group as True, i.e.:
group time C first_in_group
0 1 1 6 True
1 1 2 7 False
2 2 3 8 True
3 2 4 9 False
4 2 5 10 False
I tried several combinations of groupby, first but did not manage to achieve what I wanted.
Is there an elegant way to do it in Pandas?
Use duplicated:
df['first_in_group'] = ~df.group.duplicated()
OUTPUT:
group time C first_in_group
0 1 1 6 True
1 1 2 7 False
2 2 3 8 True
3 2 4 9 False
4 2 5 10 False
NOTE: Do the sorting 1st (if required).
df = df.sort_values(['group', 'time'])
Related
Given a frame like this
import pandas as pd
df = pd.DataFrame({'A':[1,2,3,4,6,3,7,3,2,11,13,10,1,5],'B':[1,1,1,2,2,2,2,3,3,3,3,3,4,4],
'C':[1,1,1,1,1,1,1,2,2,2,2,2,3,3]})
I want to allot a unique identifier to multiple groups in column B. For example, going from top for every two groups allot a unique identifier as shown in red boxes in below image. The end result would look like below:
Currently I am doing like below but it seems to be over kill. It's taking too much time to update even 70,000 rows:
b_unique_cnt = df['B'].nunique()
the_list = list(range(1, b_unique_cnt+1))
slice_size = 2
list_of_slices = zip(*(iter(the_list),) * slice_size)
counter = 1
df['D'] = -1
for i in list_of_slices:
df.loc[df['B'].isin(i), 'D'] = counter
counter = counter + 1
df.head(15)
You could do
df['new'] = df.B.factorize()[0]//2+1
#(df.groupby(['B'], sort=False).ngroup()//2).add(1)
df
Out[153]:
A B C new
0 1 1 1 1
1 2 1 1 1
2 3 1 1 1
3 4 2 1 1
4 6 2 1 1
5 3 2 1 1
6 7 2 1 1
7 3 3 2 2
8 2 3 2 2
9 11 3 2 2
10 13 3 2 2
11 10 3 2 2
12 1 4 3 2
13 5 4 3 2
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
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]
I have got a database with a lot of columns. Some of the rows are duplicates (on a certain subset).
Now I want to find out which row duplicates which row and put them together.
For instance, let's suppose that the data frame is
id A B C
0 0 1 2 0
1 1 2 3 4
2 2 1 4 8
3 3 1 2 3
4 4 2 3 5
5 5 5 6 2
and subset is
['A','B']
I expect something like this:
id A B C
0 0 1 2 0
1 3 1 2 3
2 1 2 3 4
3 4 2 3 5
4 2 1 4 8
5 5 5 6 2
Is there any function that can help me do this?
Thanks :)
Use DataFrame.duplicated with keep=False for mask with all dupes, then flter by boolean indexing, sorting by DataFrame.sort_values and join together by concat:
L = ['A','B']
m = df.duplicated(L, keep=False)
df = pd.concat([df[m].sort_values(L), df[~m]], ignore_index=True)
print (df)
id A B C
0 0 1 2 0
1 3 1 2 3
2 1 2 3 4
3 4 2 3 5
4 2 1 4 8
5 5 5 6 2
I have a dataframe that looks like this:
A B C
1 1 8 3
2 5 4 3
3 5 8 1
and I want to count the values so to make df like this:
total
1 2
3 2
4 1
5 2
8 2
is it possible with pandas?
With np.unique -
In [332]: df
Out[332]:
A B C
1 1 8 3
2 5 4 3
3 5 8 1
In [333]: ids, c = np.unique(df.values.ravel(), return_counts=1)
In [334]: pd.DataFrame({'total':c}, index=ids)
Out[334]:
total
1 2
3 2
4 1
5 2
8 2
With pandas-series -
In [357]: pd.Series(np.ravel(df)).value_counts().sort_index()
Out[357]:
1 2
3 2
4 1
5 2
8 2
dtype: int64
You can also use stack() and groupby()
df = pd.DataFrame({'A':[1,8,3],'B':[5,4,3],'C':[5,8,1]})
print(df)
A B C
0 1 5 5
1 8 4 8
2 3 3 1
df1 = df.stack().reset_index(1)
df1.groupby(0).count()
level_1
0
1 2
3 2
4 1
5 2
8 2
Other alternative may be to use stack, followed by value_counts then, result changed to frame and finally sorting the index:
count_df = df.stack().value_counts().to_frame('total').sort_index()
count_df
Result:
total
1 2
3 2
4 1
5 2
8 2
using np.unique(, return_counts=True) and np.column_stack():
pd.DataFrame(np.column_stack(np.unique(df, return_counts=True)))
returns:
0 1
0 1 2
1 3 2
2 4 1
3 5 2
4 8 2