putting Nans in columns that aren't within start/end numbers for each row-CodePudding

I'm trying to work out a way to put a nan in columns that don't exist between 2 start/end values in another column for each row. Say I have the below dataframe:

df = pd.DataFrame({'39' : [1, np.nan, 3],
                   '40' : [2, 4, 5],
                   '41' : [3, 1, 4],
                   '42' : [2, 5, 2],
                   '43' : [1, 1, np.nan],
                   'start' : [39, 40, 41],
                   'end' : [41, 41, 43]})

    39  40  41  42  43  start  end
0  1.0   2   3   2   1     39   41
1  NaN   4   1   5   1     40   41
2  3.0   5   4   2   3     41   43

I want to put a nan in the numbered columns that aren't between the start/end column numbers (inclusive), to get the below:

    39   40  41   42  43  start  end
0  1.0  2.0   3  NaN NaN     39   41
1  NaN  4.0   1  NaN NaN     40   41
2  NaN  NaN   4  2.0 3.0     41   43

The only way I can currently think of doing this would be to iterate through the rows or columns to check if between start and end or not, but I know iterating through dataframes is bad practice. I could turn the columns into lists and iterate through those and reassign, but I'm just wondering if there is a more efficient way to achieve this?

Edit: I should note that the numerical columns are week numbers, so it is possible for them to go over a year (e.g. 51, 52, 1, 2, 3 then start could be 51, and end could be 1). I'm wondering if I need to make a list of the column numbers to keep before doing this, as using < or > won't work for this case.

An example of this:

df2 = pd.DataFrame({'51' : [1, np.nan, 3],
                   '52' : [2, 4, 5],
                   '1' : [3, 1, 4],
                   '2' : [2, 5, 2],
                   '3' : [1, 1, 3],
                   'start' : [51, 52, 52],
                   'end' : [1, 2, 1]})

    51  52  1  2  3  start  end
0  1.0   2  3  2  1     51    1
1  NaN   4  1  5  1     52    2
2  3.0   5  4  2  3     52    1

Output:

    51  52  1    2   3  start  end
0  1.0   2  3  NaN NaN     51    1
1  NaN   4  1  5.0 NaN     52    2
2  NaN   5  4  NaN NaN     52    1

CodePudding user response：

We can make use of stack and unstack here:

df = df.set_index(["start", "end"]).stack()
idx = df.index
values = idx.get_level_values(2).astype(int)
start = idx.get_level_values(0)
end = idx.get_level_values(1)

df.where((values >= start) & (values <= end)).unstack().reset_index()

   start  end   39   40   41   42  43
0     39   41  1.0  2.0  3.0  NaN NaN
1     40   41  NaN  4.0  1.0  NaN NaN
2     41   43  NaN  NaN  4.0  2.0 NaN

CodePudding user response：

Numpy solutionwith compare between start and end :

df.columns = df.columns[:-2].astype(int).tolist()   df.columns[-2:].tolist()

s = df['start'].to_numpy()
e = df['end'].to_numpy()
cols = df.columns[:-2].to_numpy()

m = (s[:, None] <= cols) & (e[:, None] >= cols)

df.iloc[:, :-2] = df.iloc[:, :-2].where(m)
print (df)
    39   40  41   42  43  start  end
0  1.0  2.0   3  NaN NaN     39   41
1  NaN  4.0   1  NaN NaN     40   41
2  NaN  NaN   4  2.0 NaN     41   43

EDIT: If not possible compare by value with cumulative sum, for e from back and test if 1 in both masks:

s = df['start'].astype(str).to_numpy()
e = df['end'].astype(str).to_numpy()

cols = df.columns[:-2].to_numpy()

m1 = np.cumsum((s[:, None] == cols), axis=1) == 1
m2 = np.cumsum((e[:, None] == cols[::-1]), axis=1)[:, ::-1] == 1

m = m1 & m2
df.iloc[:, :-2] = df.iloc[:, :-2].where(m)
print (df)
    51  52  1    2   3  start  end
0  1.0   2  3  NaN NaN     51    1
1  NaN   4  1  5.0 NaN     52    2
2  NaN   5  4  NaN NaN     52    1

CodePudding user response：

for column in ['39', '40', '41', '42', '43']:
    df[column].loc[(float(column) < df['start']) | (float(column) > df['end'])] = np.NaN

will print:

    39   40   41   42  43  start  end
0  1.0  2.0  3.0  NaN NaN     39   41
1  NaN  4.0  1.0  NaN NaN     40   41
2  NaN  NaN  4.0  2.0 NaN     41   43