如何遍历Pandas DataFrame行并按条件设置新列值？

我正在做学校项目，不要提供完整答案。我有一个名为twitter_archive的Pandas DataFrame，包含狗狗图片评分的分子（rating_numerator）和分母（rating_denominator）：单只狗时满分10分，多只狗时满分是狗的数量×10。我想把所有评分统一成10分制，比如5只狗的40/50要调整为8/10。

以下是我的示例代码，我知道第3行语法有误，但它能准确表达我的需求：

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

我的逻辑是：检查分母是否大于10且能被10整除，若是则计算分母是10的多少倍，再将分子除以该倍数得到新分子。目前我的问题是无法将计算得到的新值对应到对应行的新列中。

该DataFrame的头部数据如下：

{'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'}}

提示方向 #

• 遍历DataFrame行的正确方式不是直接取两列迭代，可尝试用twitter_archive[['rating_numerator','rating_denominator']].iterrows()，这样能拿到每行的索引和对应值
• 计算出new_numer后，需要通过行索引定位赋值，比如twitter_archive.loc[index, 'new_numer'] = new_numer，而非直接给整列赋值，这样才能修改对应行的值
• Pandas更推荐向量化操作替代循环，可考虑用np.where或布尔索引批量处理符合条件的行，效率更高

内容的提问来源于stack exchange，提问作者Christian Love