How to append with loop in python

Question

I have been searching for hours. I have 190 columns of pivot table to loop on my script I have this script:

corr = pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[0]], list(df_pvt.columns)], method='pearson')[['X','Y','r']]

this provide output:

                                          X  ...      r
0    CORSEC_Mainstream Media_Negative Count  ...  1.000
1    CORSEC_Mainstream Media_Negative Count  ...  0.960
2    CORSEC_Mainstream Media_Negative Count  ... -0.203
3    CORSEC_Mainstream Media_Negative Count  ... -0.446
4    CORSEC_Mainstream Media_Negative Count  ...  0.488
..                                      ...  ...    ...
179  CORSEC_Mainstream Media_Negative Count  ... -0.483
180  CORSEC_Mainstream Media_Negative Count  ... -0.487
181  CORSEC_Mainstream Media_Negative Count  ...  0.145
182  CORSEC_Mainstream Media_Negative Count  ...  0.128
183  CORSEC_Mainstream Media_Negative Count  ...  0.520

[184 rows x 3 columns]

I want to append 189 other columns to my script, but this script keep providing 2 appended variables and keep replacing until the 189th variables

for var in list(range(1,189)):
    corr_all = corr.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[var]], list(df_pvt.columns)], method='pearson')[['X','Y','r']])
    print(corr_all)

Any advice?

Edit:

Its work like this:

corr = pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[0]], list(df_pvt.columns)], method='pearson')[['X','Y','r']]
corr_1 = corr.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[1]], list(df_pvt.columns)], method='pearson')[['X','Y','r']])
corr_2 = corr_1.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[2]], list(df_pvt.columns)], method='pearson')[['X','Y','r']])

But how I loop it until the corr_189?

Chaussette · Accepted Answer · 2020-05-27 09:28:30Z

1

You can try making 189 lists of values (Pearson coefficients) for each of your 189 columns, and then concatenate the columns with " df_final " which would be the dataframe containing all the 190 columns :

corr = pd.DataFrame(corr)
df_final = pd.DataFrame()

for k in range(189):
    list_Pearson_k = 'formula to compute a list of pearson values'                    
    df_list_k = pd.DataFrame(list_Pearson_k)
    df_final = pd.concat([corr,df_list_k ], axis = 1)

edited May 27, 2020 at 9:28

answered May 27, 2020 at 7:19

Chaussette

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3 Comments

Irwan Over a year ago

Sorry is that mean I need to make 189 rows of script to get 189 lists?

Chaussette Over a year ago

Fortunately no! I meant that you can do a loop, like in my answer, and that you could concatenate the column of pearson values at each iteration like this : corr = pd.DataFrame(corr) for k in range(189): list_Pearson_k = 'formula to compute a list of pearson values' df_list_k = pd.DataFrame(list_Pearson_k) df_final = pd.concat([corr,df_list_k ], axis = 1) This way you can get a df_final with your 190 columns of pearson values. If you want I can try to implement it but only in a few hours!

Chaussette Over a year ago

I changed my answer you can check that : instead of 'formula to compute a list of pearson values' you can put " pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[2]], list(df_pvt.columns)], method='pearson')[['X','Y','r']]" but I didn't try to run it

Dhaval Taunk · Accepted Answer · 2020-05-27 08:20:43Z

0

Python list append method returns None.

Change your code to this:-

corr_all = []
for var in range(1,189):
    corr_all.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[var]], list(df_pvt.columns)], method='pearson')[['X','Y','r']])
    print(corr_all)

This should help.

edited May 27, 2020 at 8:20

answered May 27, 2020 at 7:09

Dhaval Taunk

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4 Comments

James McPherson Over a year ago

list(range(1, 189)) is a bit silly, because range() gives you an in-order iterable and then that's turned into a list which.... you iterate over. Remove the call to list() and the solution is more pythonic.

Irwan Over a year ago

@Dhaval Taunk Thanks but Its still provide 2 columns combination at the end, how to generate all combination in 1 result?

Irwan Over a year ago

its work with this, but I should make like 189 rows like this. How to loop this?

corr = pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[0]], list(df_pvt.columns)], method='pearson')[['X','Y','r']] corr_1 = corr.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[1]], list(df_pvt.columns)], method='pearson')[['X','Y','r']]) corr_2 = corr_1.append(pg.pairwise_corr(df_pvt, columns=[[df_pvt.columns[2]], list(df_pvt.columns)], method='pearson')[['X','Y','r']])

Dhaval Taunk Over a year ago

@Irwan, sorry I didn't understand your question. Can you please explain properly?

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