如何遍历 pandas dataframe 的每一行，然后有条件地在该行中设置一个新值？

Question

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'}}

Answer 1

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.