如何将循环结果存储在 dataframe 中？

Question

I have the following loop creating kpi values for each of my 5 subsets.

list = [sku_1,sku_2,sku_3,sku_4,sku_5]
for i in list:
    df = triple_exp_smooth_mul(i, slen=12, extra_periods=4, alpha=0.3,beta=0.2, phi=0.9, gamma=0.2)
    kpi(df)
    print(kpi)

Output: 
Bias: 3.19, 1.06%
MAPE: 14.93%
MAE: 43.94, 14.64%
RMSE: 56.84, 18.94%
<function kpi at 0x000002634941C8B0>
Bias: -2.57, -1.31%
MAPE: 15.27%
MAE: 28.04, 14.27%
RMSE: 40.29, 20.50%
<function kpi at 0x000002634941C8B0>
Bias: -7.93, -3.77%
MAPE: 16.98%
MAE: 33.76, 16.08%
RMSE: 48.83, 23.25%
<function kpi at 0x000002634941C8B0>
Bias: -15.05, -2.98%
MAPE: 37.16%
MAE: 170.31, 33.75%
RMSE: 224.58, 44.50%
<function kpi at 0x000002634941C8B0>
Bias: 5.37, 2.03%
MAPE: 21.45%
MAE: 53.89, 20.41%
RMSE: 78.66, 29.79%
<function kpi at 0x000002634941C8B0>

Is there a way to store this in a nice table, something like this:

KPI	SKU1 Value	SKU1 Scaled
BIAS	xx	%
MAPE	xx	%
MAE	xx	%
RMSE	xx	%

With the SKU Value and SKU scaled in the columns?

Edit: Code used to create KPI function:

#Defining the function for measuring the forecast accuracy for each Forecast "#Credits to Nicolas Candeput "Data Science For Supply Chain Forecasting""
def kpi(df):
    dem_ave = df.loc[df['Error'].notnull(),'Demand'].mean()
    bias_abs = df['Error'].mean()
    bias_rel = bias_abs / dem_ave
    print('Bias: {:0.2f}, {:.2%}'.format(bias_abs,bias_rel))
    MAPE = (df['Error'].abs()/df['Demand']).mean()
    print('MAPE: {:.2%}'.format(MAPE))
    MAE_abs = df['Error'].abs().mean()
    MAE_rel = MAE_abs / dem_ave
    print('MAE: {:0.2f}, {:.2%}'.format(MAE_abs,MAE_rel))
    RMSE_abs = np.sqrt((df['Error']**2).mean())
    RMSE_rel = RMSE_abs / dem_ave
    print('RMSE: {:0.2f}, {:.2%}'.format(RMSE_abs,RMSE_rel))

Answer 1

The first thing to understand is that printing something inside a function isn't the same as returning it. See How is returning the output of a function different from printing it?

I also second Tim's suggestion that Bias, etc. should be columns , and you should have rows for each SKU.

You should return the values you want from your function, and then use those returned values to add a row to your dataframe. We will return the values as a dictionary, save all returned dicts to a list, and once we're done with all SKUs, create a dataframe from this list of dicts using the constructor (see Convert list of dictionaries to a pandas DataFrame ).

So first, let's modify our kpi() function to return a dict whose keys are the column names, and values are the values we want for that row.

def kpi(df):
    ret = {}
    dem_ave = df.loc[df['Error'].notnull(),'Demand'].mean()
    bias_abs = df['Error'].mean()
    bias_rel = bias_abs / dem_ave
    ret['Bias'] = bias_abs
    ret['Bias %'] = bias_rel
    # print('Bias: {:0.2f}, {:.2%}'.format(bias_abs,bias_rel))
    MAPE = (df['Error'].abs()/df['Demand']).mean()
    ret['MAPE'] = MAPE
    # print('MAPE: {:.2%}'.format(MAPE))
    MAE_abs = df['Error'].abs().mean()
    MAE_rel = MAE_abs / dem_ave
    ret['MAE'] = MAE_abs
    ret['MAE %'] = MAE_rel
    # print('MAE: {:0.2f}, {:.2%}'.format(MAE_abs,MAE_rel))
    RMSE_abs = np.sqrt((df['Error']**2).mean())
    RMSE_rel = RMSE_abs / dem_ave
    ret['RMSE'] = RMSE_abs
    ret['RMSE %'] = RMSE_rel
    # print('RMSE: {:0.2f}, {:.2%}'.format(RMSE_abs,RMSE_rel))
    return ret

Now, let's create an empty list to hold our returned dicts and fill it up:

sku_list = [sku_1,sku_2,sku_3,sku_4,sku_5]
result_list = []
for sku in sku_list:
    df = triple_exp_smooth_mul(sku, slen=12, extra_periods=4, alpha=0.3,beta=0.2, phi=0.9, gamma=0.2)
    result_list.append(kpi(df))

You could write this as a list comprehension instead of a loop like so:

result_list = [kpi(
                   triple_exp_smooth_mul(sku, slen=12, extra_periods=4, alpha=0.3,beta=0.2, phi=0.9, gamma=0.2)
                   ) 
               for sku in sku_list]

Then, create a dataframe out of it:

result_df = pd.DataFrame(result_list)

Note that I renamed your list to sku_list because it's not a good idea to shadow builtins .