使用 .iterrows() 和 series.nlargest() 获得 Dataframe 中连续的最高数字

Question

I am trying to create a function that uses df.iterrows() and Series.nlargest . I want to iterate over each row and find the largest number and then mark it as a 1 . This is the data frame:

A   B    C
9   6    5
3   7    2

Here is the output I wish to have:

A    B   C
1    0   0
0    1   0

This is the function I wish to use here:

def get_top_n(df, top_n):
    """


    Parameters
    ----------
    df : DataFrame

    top_n : int
        The top number to get
    Returns
    -------
    top_numbers : DataFrame
    Returns the top number marked with a 1

    """
    # Implement Function
    for row in df.iterrows():
        top_numbers = row.nlargest(top_n).sum()

    return top_numbers

I get the following error: AttributeError: 'tuple' object has no attribute 'nlargest'

Help would be appreciated on how to re-write my function in a neater way and to actually work! Thanks in advance

Answer 1

Add i variable, because iterrows return indices with Series for each row:

for i, row in df.iterrows():
    top_numbers = row.nlargest(top_n).sum()

General solution with numpy.argsort for positions in descending order , then compare and convert boolean array to integers:

def get_top_n(df, top_n):
    if top_n > len(df.columns):
        raise ValueError("Value is higher as number of columns")
    elif not isinstance(top_n, int):
        raise ValueError("Value is not integer")

    else:
        arr = ((-df.values).argsort(axis=1) < top_n).astype(int)
        df1 = pd.DataFrame(arr, index=df.index, columns=df.columns)
        return (df1)

df1 = get_top_n(df, 2)
print (df1)
   A  B  C
0  1  1  0
1  1  1  0

df1 = get_top_n(df, 1)
print (df1)
   A  B  C
0  1  0  0
1  0  1  0

EDIT:

Solution with iterrows is possible, but not recommended, because slow:

top_n = 2
for i, row in df.iterrows():
    top = row.nlargest(top_n).index
    df.loc[i] = 0
    df.loc[i, top] = 1

print (df)
   A  B  C
0  1  1  0
1  1  1  0

Answer 2

For context, the dataframe consists of stock return data for the S&P500 over approximately 4 years

def get_top_n(prev_returns, top_n):

    # generate dataframe populated with zeros for merging
    top_stocks = pd.DataFrame(0, columns = prev_returns.columns, index = prev_returns.index)

    # find top_n largest entries by row
    df = prev_returns.apply(lambda x: x.nlargest(top_n), axis=1)

    # merge dataframes
    top_stocks = top_stocks.merge(df, how = 'right').set_index(df.index)

    # return dataframe replacing non_zero answers with a 1
    return (top_stocks.notnull()) * 1

Answer 3

Alternatively, the 2-line solution could be

---

def get_top_n(df, top_n):

    # find top_n largest entries by stock
    df = df.apply(lambda x: x.nlargest(top_n), axis=1)

    # convert dataframe NaN or float entries True and False, and then convert to 0 and 1
    top_numbers = (df.notnull()).astype(np.int)

    return top_numbers