我需要根据这些条件更新列值
i. if score > 3, set score to 1.
ii. if score <= 2, set score to 0.
iii. if score == 3, drop that row.
分数的取值范围是1到5
我编写了以下代码,但是所有值都被更改为0。
reviews.loc[reviews['Score'] > 3, 'Score'] = 1
reviews.loc[reviews['Score'] <= 2, 'Score'] = 0
reviews.drop(reviews[reviews['Score'] == 3].index, inplace = True)
请指出错误在于此。
答案 0 :(得分:2)
存在逻辑问题:
reviews = pd.DataFrame({'Score':range(6)})
print (reviews)
Score
0 0
1 1
2 2
3 3
4 4
5 5
如果将所有值都设置得较高,例如3
到1
,则它的工作原理如下:
reviews.loc[reviews['Score'] > 3, 'Score'] = 1
print (reviews)
Score
0 0
1 1
2 2
3 3
4 1
5 1
然后将所有没有3
的vallus设置为0
,因此也会从1
中替换reviews['Score'] > 3
:
reviews.loc[reviews['Score'] <= 2, 'Score'] = 0
print (reviews)
Score
0 0
1 0
2 0
3 3
4 0
5 0
最后删除的3
行仅获得0
值:
reviews.drop(reviews[reviews['Score'] == 3].index, inplace = True)
print (reviews)
Score
0 0
1 0
2 0
4 0
5 0
您可以更改解决方案:
reviews = pd.DataFrame({'Score':range(6)})
print (reviews)
Score
0 0
1 1
2 2
3 3
4 4
5 5
首先通过过滤boolean indexing
中所有不等于3
的行来删除3
:
reviews = reviews[reviews['Score'] != 3].copy()
然后将值设置为0
和1
:
reviews['Score'] = (reviews['Score'] > 3).astype(int)
#alternative
reviews['Score'] = np.where(reviews['Score'] > 3, 1, 0)
print (reviews)
Score
0 0
1 0
2 0
4 1
5 1
EDIT1:
您应该使用交换行来更改您的解决方案-首先设置0
,然后再设置1
,以避免覆盖值:
reviews.loc[reviews['Score'] <= 2, 'Score'] = 0
reviews.loc[reviews['Score'] > 3, 'Score'] = 1
reviews.drop(reviews[reviews['Score'] == 3].index, inplace = True)
print (reviews)
Score
0 0
1 0
2 0
4 1
5 1