1

I am working on a school project, so please no exact answers. I have a pandas dataframe that has numerators and denominators rating images of dogs out of 10. When there are multiple dogs in the image, the rating is out of number of dogs * 10. I am trying to adjust it so that for example... if there are 5 dogs, and the rating is 40/50, then the new numerator/denominator is 8/10. Here is an example of my code. I am aware that the syntax does not work in line 3, but I believe it accurately represents what I am trying to accomplish. twitter_archive is the dataframe.

twitter_archive['new_denom'] = 10
twitter_archive['new_numer'] = 0
for numer, denom in twitter_archive['rating_numerator','rating_denominator']:
    if (denom > 10) & (denom % 10 == 0):
        num_denom = denom / 10
        new_numer = numer / num_denom
        twitter_archive['new_numer'] = new_numer

So basically I am checking the denominator if it is above 10, and if it is, is it divisible by 10? if it is, then find out how many times 10 goes into it, and then divide the numerator by that value to get an new numerator. I think my logic for that works fine, but the issue I have is grabbing that row, and then adding that new value to the new column I created, in that row. edit: added df head

tweet_id timestamp text rating_numerator rating_denominator name doggo floofer pupper puppo avg_numerator avg_denom avg_numer
0 8.924206e+17 2017-08-01 16:23:56+00:00 This is Phineas. He's a mystical boy. Only eve... 13.0 10.0 phineas None None None None 0.0 10 0
1 8.921774e+17 2017-08-01 00:17:27+00:00 This is Tilly. She's just checking pup on you.... 13.0 10.0 tilly None None None None 0.0 10 0
2 8.918152e+17 2017-07-31 00:18:03+00:00 This is Archie. He is a rare Norwegian Pouncin... 12.0 10.0 archie None None None None 0.0 10 0
3 8.916896e+17 2017-07-30 15:58:51+00:00 This is Darla. She commenced a snooze mid meal... 13.0 10.0 darla None None None None 0.0 10 0
4 8.913276e+17 2017-07-29 16:00:24+00:00 This is Franklin. He would like you to stop ca... 12.0 10.0 franklin None None None None 0.0 10 0

copy/paste head below:

{'tweet_id': {0: 8.924206435553362e+17,
  1: 8.921774213063434e+17,
  2: 8.918151813780849e+17,
  3: 8.916895572798587e+17,
  4: 8.913275589266883e+17},
 'timestamp': {0: Timestamp('2017-08-01 16:23:56+0000', tz='UTC'),
  1: Timestamp('2017-08-01 00:17:27+0000', tz='UTC'),
  2: Timestamp('2017-07-31 00:18:03+0000', tz='UTC'),
  3: Timestamp('2017-07-30 15:58:51+0000', tz='UTC'),
  4: Timestamp('2017-07-29 16:00:24+0000', tz='UTC')},
 'text': {0: "This is Phineas. He's a mystical boy. Only ever appears in the hole of a donut. 13/10 ",
  1: "This is Tilly. She's just checking pup on you. Hopes you're doing ok. If not, she's available for pats, snugs, boops, the whole bit. 13/10 ",
  2: 'This is Archie. He is a rare Norwegian Pouncing Corgo. Lives in the tall grass. You never know when one may strike. 12/10 ',
  3: 'This is Darla. She commenced a snooze mid meal. 13/10 happens to the best of us ',
  4: 'This is Franklin. He would like you to stop calling him "cute." He is a very fierce shark and should be respected as such. 12/10 #BarkWeek '},
 'rating_numerator': {0: 13.0, 1: 13.0, 2: 12.0, 3: 13.0, 4: 12.0},
 'rating_denominator': {0: 10.0, 1: 10.0, 2: 10.0, 3: 10.0, 4: 10.0},
 'name': {0: 'phineas', 1: 'tilly', 2: 'archie', 3: 'darla', 4: 'franklin'},
 'doggo': {0: 'None', 1: 'None', 2: 'None', 3: 'None', 4: 'None'},
 'floofer': {0: 'None', 1: 'None', 2: 'None', 3: 'None', 4: 'None'},
 'pupper': {0: 'None', 1: 'None', 2: 'None', 3: 'None', 4: 'None'},
 'puppo': {0: 'None', 1: 'None', 2: 'None', 3: 'None', 4: 'None'}}
Christian Love
  • 49
  • 6

1 Answers1

2

If you want to use for loop to get row values, you can use iterrows() function.

for idx, row in twitter_archive.iterrows():
    denom = row['rating_denominator']
    numer = row['rating_numerator']
    # You can add values in list and concat it with df

Faster way to iterate on df is itertuples():

for row in twitter_archive.itertuples():
    denom = row[1]
    numer = row[2]

But I think best way to create new col from old ones is to use pandas apply function .

df = pd.DataFrame(data={'a' : [1,2], 'b': [3,5]})
df['c'] = df.apply(lambda x: 'sum_is_odd' if (x['a'] + x['b']) % 2 == 1 else 'sum_is_even', axis=1)

In this case, 'c' is a new column and value is calculated using 'a' and 'b' columns.

Zorojuro
  • 71
  • 5
  • Thanks for your help. How would I use another column from my dataframe inside of the lambda? for example I would need to do something like this I think... ```twitter_archive['new_numer'] = twitter_archive['rating_numerator'].apply(lambda x: (twitter_archive['rating_numerator'] / (twitter_archive['rating_denominator'] / 10)) if ((twitter_archive['rating_denominator'] > 10) & (twitter_archive['rating_denominator'] % 10 == 0)) else twitter_archive['rating_numerator'], axis=1) ``` but I get this error: TypeError: () got an unexpected keyword argument 'axis' – Christian Love Jul 25 '22 at 20:57
  • Can you elaborate on what you mean by "use x like row"? I noticed a mistake I made near .apply which was putting the column name on the dataframe, but I still get an error (with and without axis=1) : ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all(). – Christian Love Jul 25 '22 at 21:13
  • 1
    I add example in answer. Please tell me if it is helpful. – Zorojuro Jul 25 '22 at 21:16
  • This did solve my problem. Thank you for the help! I ended up using the apply function. Cheers! – Christian Love Jul 25 '22 at 21:39
  • Is is possible to set two true values? for example, if this is true, set x['a'] = 1 AND x['b'] = 2? or will it just need to be done in multiple .apply functions? – Christian Love Jul 25 '22 at 21:43
  • 1
    Take a look https://stackoverflow.com/questions/12356501/pandas-create-two-new-columns-in-a-dataframe-with-values-calculated-from-a-pre – Zorojuro Jul 25 '22 at 21:46