Is it possible to obtain groupby style counts without collapsing Pandas DataFrame? - python

I have a DataFrame with 9 columns, and I'm trying to add a column of counts of unique values based on the first 3 columns (e.g. Cols A, B, and C, must match to count as a unique value , but the remaining columns can vary. I attempted to do this as with groupby:
df = pd.DataFrame(resultsFile500.groupby(['chr','start','end']).size().reset_index().rename(columns={0:'count'}))
This returns a DataFrame with 5 columns, and the counts are what I want. However, I also need values from the original data frame, so what I have been trying to do is somehow get those values of counts as a column in the original df. So, this would mean that if two rows in columns chr, start, and end, had identical values, the counts column would be 2 in both rows, but they would not be collapsed to one row. Is there an easy solution here that I'm missing, or do I need to hack something together?

You can use .transform to get non-collapsing behavior:
>>> df
a b c d e
0 3 4 1 3 0
1 3 1 4 3 0
2 4 3 3 2 1
3 3 4 1 4 0
4 0 4 3 3 2
5 1 2 0 4 1
6 3 1 4 2 1
7 0 4 3 4 0
8 1 3 0 1 1
9 3 4 1 2 1
>>> df.groupby(['a','b','c']).transform('count')
d e
0 3 3
1 2 2
2 1 1
3 3 3
4 2 2
5 1 1
6 2 2
7 2 2
8 1 1
9 3 3
>>>
Note, i'll have to choose an arbitrary column from the .transform result, but then just do:
>>> df['unique_count'] = df.groupby(['a','b','c']).transform('count')['d']
>>> df
a b c d e unique_count
0 3 4 1 3 0 3
1 3 1 4 3 0 2
2 4 3 3 2 1 1
3 3 4 1 4 0 3
4 0 4 3 3 2 2
5 1 2 0 4 1 1
6 3 1 4 2 1 2
7 0 4 3 4 0 2
8 1 3 0 1 1 1
9 3 4 1 2 1 3

Related

How to identify one column with continuous number and same value of another column?

I have a DataFrame with two columns A and B.
I want to create a new column named C to identify the continuous A with the same B value.
Here's an example
import pandas as pd
df = pd.DataFrame({'A':[1,2,3,5,6,10,11,12,13,18], 'B':[1,1,2,2,3,3,3,3,4,4]})
I found a similar question, but that method only identifies the continuous A regardless of B.
df['C'] = df['A'].diff().ne(1).cumsum().sub(1)
I have tried to groupby B and apply the function like this:
df['C'] = df.groupby('B').apply(lambda x: x['A'].diff().ne(1).cumsum().sub(1))
However, it doesn't work: TypeError: incompatible index of inserted column with frame index.
The expected output is
A B C
1 1 0
2 1 0
3 2 1
5 2 2
6 3 3
10 3 4
11 3 4
12 3 4
13 4 5
18 4 6
Let's create a sequential counter using groupby, diff and cumsum then factorize to reencode the counter
df['C'] = df.groupby('B')['A'].diff().ne(1).cumsum().factorize()[0]
Result
A B C
0 1 1 0
1 2 1 0
2 3 2 1
3 5 2 2
4 6 3 3
5 10 3 4
6 11 3 4
7 12 3 4
8 13 4 5
9 18 4 6
Use DataFrameGroupBy.diff with compare not equal 1 and Series.cumsum, last subtract 1:
df['C'] = df.groupby('B')['A'].diff().ne(1).cumsum().sub(1)
print (df)
A B C
0 1 1 0
1 2 1 0
2 3 2 1
3 5 2 2
4 6 3 3
5 10 3 4
6 11 3 4
7 12 3 4
8 13 4 5
9 18 4 6

Count the most frequent values in a row pandas and make a column with that most frequent value

I have a data frame like this below:
a b c
0 3 3 3
1 3 3 3
2 3 3 3
3 3 3 3
4 2 3 2
5 3 3 3
6 1 2 1
7 2 3 2
8 0 0 0
9 0 1 0
I want to count frequency of each row and add a column result containing the max frequency like this below:
a b c result
0 3 3 3 3
1 3 3 3 3
2 3 3 3 3
3 3 3 3 3
4 2 3 2 2
5 3 3 3 3
6 1 2 1 1
7 2 3 2 2
8 0 0 0 0
9 0 1 0 0
I tries to do transpose and looping through the transposed columns to get the value_counts but could not got the right result.
Any help is highly appreciated.
Use DataFrame.mode with select first column by positions with DataFrame.iloc:
df['result'] = df.mode(axis=1).iloc[:, 0]
print (df)
a b c result
0 3 3 3 3
1 3 3 3 3
2 3 3 3 3
3 3 3 3 3
4 2 3 2 2
5 3 3 3 3
6 1 2 1 1
7 2 3 2 2
8 0 0 0 0
9 0 1 0 0

Padding and reshaping pandas dataframe

I have a dataframe with the following form:
data = pd.DataFrame({'ID':[1,1,1,2,2,2,2,3,3],'Time':[0,1,2,0,1,2,3,0,1],
'sig':[2,3,1,4,2,0,2,3,5],'sig2':[9,2,8,0,4,5,1,1,0],
'group':['A','A','A','B','B','B','B','A','A']})
print(data)
ID Time sig sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 2 0 4 0 B
4 2 1 2 4 B
5 2 2 0 5 B
6 2 3 2 1 B
7 3 0 3 1 A
8 3 1 5 0 A
I want to reshape and pad such that each 'ID' has the same number of Time values, the sig1,sig2 are padded with zeros (or mean value within ID) and the group carries the same letter value. The output after repadding would be :
data_pad = pd.DataFrame({'ID':[1,1,1,1,2,2,2,2,3,3,3,3],'Time':[0,1,2,3,0,1,2,3,0,1,2,3],
'sig1':[2,3,1,0,4,2,0,2,3,5,0,0],'sig2':[9,2,8,0,0,4,5,1,1,0,0,0],
'group':['A','A','A','A','B','B','B','B','A','A','A','A']})
print(data_pad)
ID Time sig1 sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 1 3 0 0 A
4 2 0 4 0 B
5 2 1 2 4 B
6 2 2 0 5 B
7 2 3 2 1 B
8 3 0 3 1 A
9 3 1 5 0 A
10 3 2 0 0 A
11 3 3 0 0 A
My end goal is to ultimately reshape this into something with shape (number of ID, number of time points, number of sequences {2 here}).
It seems that if I pivot data, it fills in with nan values, which is fine for the signal values, but not the groups. I am also hoping to avoid looping through data.groupby('ID'), since my actual data has a large number of groups and the looping would likely be very slow.
Here's one approach creating the new index with pd.MultiIndex.from_product and using it to reindex on the Time column:
df = data.set_index(['ID', 'Time'])
# define a the new index
ix = pd.MultiIndex.from_product([df.index.levels[0],
df.index.levels[1]],
names=['ID', 'Time'])
# reindex using the above multiindex
df = df.reindex(ix, fill_value=0)
# forward fill the missing values in group
df['group'] = df.group.mask(df.group.eq(0)).ffill()
print(df.reset_index())
ID Time sig sig2 group
0 1 0 2 9 A
1 1 1 3 2 A
2 1 2 1 8 A
3 1 3 0 0 A
4 2 0 4 0 B
5 2 1 2 4 B
6 2 2 0 5 B
7 2 3 2 1 B
8 3 0 3 1 A
9 3 1 5 0 A
10 3 2 0 0 A
11 3 3 0 0 A
IIUC:
(data.pivot_table(columns='Time', index=['ID','group'], fill_value=0)
.stack('Time')
.sort_index(level=['ID','Time'])
.reset_index()
)
Output:
ID group Time sig sig2
0 1 A 0 2 9
1 1 A 1 3 2
2 1 A 2 1 8
3 1 A 3 0 0
4 2 B 0 4 0
5 2 B 1 2 4
6 2 B 2 0 5
7 2 B 3 2 1
8 3 A 0 3 1
9 3 A 1 5 0
10 3 A 2 0 0
11 3 A 3 0 0

Python Counting Same Values For Specific Columns

If i have a dataframe;
A B C D
1 1 2 2 1
2 1 1 2 1
3 3 1 0 1
4 2 4 4 4
I want to make addition B and C columns and counting whether or not the same values with D columns. Desired output is;
A B C B+C D
1 1 2 2 4 1
2 1 1 2 3 1
3 3 1 0 1 1
4 2 4 4 8 4
There are 3 different values compare the "B+C" and "D".
Could you please help me about this?
You could do something like:
df.B.add(df.C).ne(df.D).sum()
# 3
If you need to add the column:
df['B+C'] = df.B.add(df.C)
diff = df['B+C'].ne(df.D).sum()
print(f'There are {diff} different values compare the "B+C" and "D"')
#There are 3 different values compare the "B+C" and "D"
df.insert(3,'B+C', df['B']+df['C'])
3 is the index
df.head()
A B C B+C D
0 1 2 2 4 1
1 1 1 2 3 1
2 3 1 0 1 1
3 2 4 4 8 4
After that you can follow the steps of #yatu
df['B+C'].ne(df['D'])
0 True
1 True
2 False
3 True dtype: bool
df['B+C'].ne(df['D']).sum()
3

Pandas reverse column values groupwise

I want to reverse a column values in my dataframe, but only on a individual "groupby" level. Below you can find a minimal demonstration example, where I want to "flip" values that belong the same letter A,B or C:
df = pd.DataFrame({"group":["A","A","A","B","B","B","B","C","C"],
"value": [1,3,2,4,4,2,3,2,5]})
group value
0 A 1
1 A 3
2 A 2
3 B 4
4 B 4
5 B 2
6 B 3
7 C 2
8 C 5
My desired output looks like this: (column is added instead of replaced only for the brevity purposes)
group value value_desired
0 A 1 2
1 A 3 3
2 A 2 1
3 B 4 3
4 B 4 2
5 B 2 4
6 B 3 4
7 C 2 5
8 C 5 2
As always, when I don't see a proper vector-style approach, I end messing with loops just for the sake of final output, but my current code hurts me very much:
for i in list(set(df["group"].values.tolist())):
reversed_group = df.loc[df["group"]==i,"value"].values.tolist()[::-1]
df.loc[df["group"]==i,"value_desired"] = reversed_group
Pandas gurus, please show me the way :)
You can use transform
In [900]: df.groupby('group')['value'].transform(lambda x: x[::-1])
Out[900]:
0 2
1 3
2 1
3 3
4 2
5 4
6 4
7 5
8 2
Name: value, dtype: int64
Details
In [901]: df['value_desired'] = df.groupby('group')['value'].transform(lambda x: x[::-1])
In [902]: df
Out[902]:
group value value_desired
0 A 1 2
1 A 3 3
2 A 2 1
3 B 4 3
4 B 4 2
5 B 2 4
6 B 3 4
7 C 2 5
8 C 5 2

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