I have a list of index numbers that represent index locations for a DF. list_index = [2,7,12]
I want to sum from a single column in the DF by rolling through each number in list_index and totaling the counts between the index points (and restart count at 0 at each index point). Here is a mini example.
The desired output is in OUTPUT column, which increments every time there is another 1 from COL 1 and RESTARTS the count at 0 on the location after the number in the list_index.
I was able to get it to work with a loop but there are millions of rows in the DF and it takes a while for the loop to run. It seems like I need a lambda function with a sum but I need to input start and end point in index.
Something like lambda x:x.rolling(start_index, end_index).sum()? Can anyone help me out on this.
You can try of cummulative sum and retrieving only 1 values related information , rolling sum with diffferent intervals is not possible
a = df['col'].eq(1).cumsum()
df['output'] = a - a.mask(df['col'].eq(1)).ffill().fillna(0).astype(int)
Out:
col output
0 0 0
1 1 1
2 1 2
3 0 0
4 1 1
5 1 2
6 1 3
7 0 0
8 0 0
9 0 0
10 0 0
11 1 1
12 1 2
13 0 0
14 0 0
15 1 1
Related
df = pd.DataFrame({'id': ['id1', 'id1','id1', 'id2','id1','id1','id1'],
'activity':['swimming','running','jogging','walking','walking','walking','walking'],
'month':[2,3,4,3,4,4,3]})
pd.crosstab(df['id'], df['activity'])
I'd like to add another column for month in the output to get counts per user within each month for the respective activity.
df.set_index(['id','month'])['activity'].unstack().reset_index()
I get error.
edit: Expected output in the image. I do not know how to create a table.
You can pass a list of columns to pd.crosstab:
x = pd.crosstab([df["id"], df["month"]], df["activity"]).reset_index()
x.columns.name = None
print(x)
Prints:
id month jogging running swimming walking
0 id1 2 0 0 1 0
1 id1 3 0 1 0 1
2 id1 4 1 0 0 2
3 id2 3 0 0 0 1
I have data that looks like this:
CHROM POS REF ALT ... is_sever_int is_sever_str is_sever_f encoding_str
0 chr1 14907 A G ... 1 1 one one
1 chr1 14930 A G ... 1 1 one one
These are the columns that I'm interested to perform calculations on (example) :
is_severe snp _id encoding
1 1 one
1 1 two
0 1 one
1 2 two
0 2 two
0 2 one
what I want to do is to count for each snp_id and severe_id how many ones and twos are in the encoding column :
snp_id is_svere encoding_one encoding_two
1 1 1 1
1 0 1 0
2 1 0 1
2 0 1 1
I tried this :
df.groupby(["snp_id","is_sever_f","encoding_str"])["encoding_str"].count()
but it gave the error :
incompatible index of inserted column with frame index
then i tried this:
df["count"]=df.groupby(["snp_id","is_sever_f","encoding_str"],as_index=False)["encoding_str"].count()
and it returned:
Expected a 1D array, got an array with shape (2532831, 3)
how can i fix this? thank you:)
Let's try groupby with whole columns and get size of each group then unstack the encoding index.
out = (df.groupby(['is_severe', 'snp_id', 'encoding']).size()
.unstack(fill_value=0)
.add_prefix('encoding_')
.reset_index())
print(out)
encoding is_severe snp_id encoding_one encoding_two
0 0 1 1 0
1 0 2 1 1
2 1 1 1 1
3 1 2 0 1
Try as follows:
Use pd.get_dummies to convert categorical data in column encoding into indicator variables.
Chain df.groupby and get sum to turn double rows per group into one row (i.e. [0,1] and [1,0] will become [1,1] where df.snp_id == 2 and df.is_severe == 0).
res = pd.get_dummies(data=df, columns=['encoding'])\
.groupby(['snp_id','is_severe'], as_index=False, sort=False).sum()
print(res)
snp_id is_severe encoding_one encoding_two
0 1 1 1 1
1 1 0 1 0
2 2 1 0 1
3 2 0 1 1
If your actual df has more columns, limit the assigment to the data parameter inside get_dummies. I.e. use:
res = pd.get_dummies(data=df[['is_severe', 'snp_id', 'encoding']],
columns=['encoding']).groupby(['snp_id','is_severe'],
as_index=False, sort=False)\
.sum()
I have the following dataframe:
d = {'value': [1,1,1,1,1,1,1,1,1,1], 'flag_1': [0,1,0,1,1,1,0,1,1,1],'flag_2':[1,0,1,1,1,1,1,0,1,1],'index':[1,2,3,4,5,6,7,8,9,10]}
df = pd.DataFrame(data=d)
I need to perform the following filter on it:
If flag 1 and flag 2 are equal keep the row with the maximum index from the consecutive indices. Below for rows 4,5,6 and rows 9,10 flag 1 and flag 2 are equal. From the group of consecutive indices 4,5,6 therefore I wish to keep only row 6 and drop rows 4 and 5. For the next group of rows 9 and 10 I wish to keep only row 10. The rows where flag 1 and 2 are not equal should all be retained. I want my final output to look as shown below:
I am really not sure how to achieve what is required so I would be grateful for any advice on how to do it.
IIUC, you can compare consecutive rows with shift. This solution requires a sorted index.
In [5]: df[~df[['flag_1', 'flag_2']].eq(df[['flag_1', 'flag_2']].shift(-1)).all(axis=1)]
Out[5]:
value flag_1 flag_2 index
0 1 0 1 1
1 1 1 0 2
2 1 0 1 3
5 1 1 1 6
6 1 0 1 7
7 1 1 0 8
9 1 1 1 10
I'm using groupby on a pandas dataframe to drop all rows that don't have the minimum of a specific column. Something like this:
df1 = df.groupby("item", as_index=False)["diff"].min()
However, if I have more than those two columns, the other columns (e.g. otherstuff in my example) get dropped. Can I keep those columns using groupby, or am I going to have to find a different way to drop the rows?
My data looks like:
item diff otherstuff
0 1 2 1
1 1 1 2
2 1 3 7
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
and should end up like:
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
but what I'm getting is:
item diff
0 1 1
1 2 -6
2 3 0
I've been looking through the documentation and can't find anything. I tried:
df1 = df.groupby(["item", "otherstuff"], as_index=false)["diff"].min()
df1 = df.groupby("item", as_index=false)["diff"].min()["otherstuff"]
df1 = df.groupby("item", as_index=false)["otherstuff", "diff"].min()
But none of those work (I realized with the last one that the syntax is meant for aggregating after a group is created).
Method #1: use idxmin() to get the indices of the elements of minimum diff, and then select those:
>>> df.loc[df.groupby("item")["diff"].idxmin()]
item diff otherstuff
1 1 1 2
6 2 -6 2
7 3 0 0
[3 rows x 3 columns]
Method #2: sort by diff, and then take the first element in each item group:
>>> df.sort_values("diff").groupby("item", as_index=False).first()
item diff otherstuff
0 1 1 2
1 2 -6 2
2 3 0 0
[3 rows x 3 columns]
Note that the resulting indices are different even though the row content is the same.
You can use DataFrame.sort_values with DataFrame.drop_duplicates:
df = df.sort_values(by='diff').drop_duplicates(subset='item')
print (df)
item diff otherstuff
6 2 -6 2
7 3 0 0
1 1 1 2
If possible multiple minimal values per groups and want all min rows use boolean indexing with transform for minimal values per groups:
print (df)
item diff otherstuff
0 1 2 1
1 1 1 2 <-multiple min
2 1 1 7 <-multiple min
3 2 -1 0
4 2 1 3
5 2 4 9
6 2 -6 2
7 3 0 0
8 3 2 9
print (df.groupby("item")["diff"].transform('min'))
0 1
1 1
2 1
3 -6
4 -6
5 -6
6 -6
7 0
8 0
Name: diff, dtype: int64
df = df[df.groupby("item")["diff"].transform('min') == df['diff']]
print (df)
item diff otherstuff
1 1 1 2
2 1 1 7
6 2 -6 2
7 3 0 0
The above answer worked great if there is / you want one min. In my case there could be multiple mins and I wanted all rows equal to min which .idxmin() doesn't give you. This worked
def filter_group(dfg, col):
return dfg[dfg[col] == dfg[col].min()]
df = pd.DataFrame({'g': ['a'] * 6 + ['b'] * 6, 'v1': (list(range(3)) + list(range(3))) * 2, 'v2': range(12)})
df.groupby('g',group_keys=False).apply(lambda x: filter_group(x,'v1'))
As an aside, .filter() is also relevant to this question but didn't work for me.
I tried everyone's method and I couldn't get it to work properly. Instead I did the process step-by-step and ended up with the correct result.
df.sort_values(by='item', inplace=True, ignore_index=True)
df.drop_duplicates(subset='diff', inplace=True, ignore_index=True)
df.sort_values(by=['diff'], inplace=True, ignore_index=True)
For a little more explanation:
Sort items by the minimum value you want
Drop the duplicates of the column you want to sort with
Resort the data because the data is still sorted by the minimum values
If you know that all of your "items" have more than one record you can sort, then use duplicated:
df.sort_values(by='diff').duplicated(subset='item', keep='first')
I am creating an excel sheet that has three columns. Detail, month and month count
1 -- I would like for the formula to look at the detail column and if there is text add the previous cell number plus 1 to new month count, if not insert 0
2-- I would like the formula to add the previous cell before the cell with 0 and for the cell with 0 not to impact the other cells or reset the cells back to 1 witch is the problem am having
3-- I also need the formula to reset for every month from what ever number it was back to 0 or 1 depending if the new month first cell has text or not. for this I need the formula to look at the month column
This is what I have so far:
=IF(ISTEXT(G95), I94+ 1, 0)
The formula for the count column should be as follows.
=IF(A2<>"",COUNTIF($B$1:B2,B2)-COUNTIFS($A$1:A2,"",$B$1:B2,B2),0)
Breakdown of how this works:
A2<>"" Will check if the detail column is populated
COUNTIF($B$1:B2,B2) will figure out how many entries are above this row that reference the same month.
COUNTIFS($A$1:A2,"",$B$1:B2,B2) Will find how many cells are blank provided that it also matches the month. This subtracted from the previous section gives you how many are not blank.
The IF will return 0 if the detail is empty.
Which returned the following data
Orderly Random
Det Mon Count Det Mon Count
X 1 1 2 0
X 1 2 X 1 1
X 1 3 X 1 2
1 0 2 0
X 1 4 X 2 1
X 2 1 X 1 3
X 2 2 X 1 4
2 0 1 0
2 0 1 0
2 0 2 0
3 0 3 0
X 3 1 X 3 1
3 0 1 0
X 3 2 3 0
X 3 3 X 1 5
3 0 X 2 2
X 3 4 X 3 2
3 0 3 0
X 3 5 3 0
X 3 6 2 0
It sounds like you want to keep a running total for the month count in the column and put a 0 if there is not text. If that is the case, you can put this formula in I95.
=IF(ISTEXT(G95),MAX($I$2:I94)+1, 0)