问题描述
Using a single class:
<__main__.MyIterator object at 0x7f65a23b6fd0>
1
2
3
Separating the iterator class from the container class:
<__main__.Iterator object at 0x7f65a23b6ca0>
1
2
3
我想遍历每一行,检查“TOTAL_COVERAGE”的值,如果是“是”,则对其他值执行数学运算,即:
SELECT
*
FROM
customer as c
INNER JOIN
salesman as s
ON c.salesman_id = s.salesman_id;
但我收到错误:df = pd.DataFrame({'ID' : ['O60829','O60341','Q9H1R3'],'TOTAL_COVERAGE' : ['yes','yes','no'],'BEG_D' : ['1','1','500'],'END_D' : ['102','25','600'],'BEG_S' : ['1','1'],'END_S': ['102','458']})
必须有一个我没有看到的简单修复。提前致谢!
解决方法
你可以不用 iterrows
和 apply
,直接等价:
df['%'] = ''
df.loc[df['TOTAL_COVERAGE'] == 'yes','%'] =
df['END_S'].astype(int) * 100 / df['END_D'].astype(int)
,
您可以用矢量化方法解决它,不需要 iterrows
和 apply
:
df['%'] = (df['END_S'].astype(int) * 100 / df['END_D'].astype(int)) \
.where(df['TOTAL_COVERAGE'] == 'yes')
df
# ID TOTAL_COVERAGE BEG_D END_D BEG_S END_S %
#0 O60829 yes 1 102 1 102 100.0
#1 O60341 yes 1 25 1 25 100.0
#2 Q9H1R3 no 500 600 1 458 NaN
您得到 keyError 的原因是因为当您使用 apply
时,lambda x
的参数是一列(pandas 系列),不能用于访问特定列顾名思义。
无需执行 iterrows()
。可以使用 numpy.where()
完成条件逻辑以提供更有效的解决方案
df = pd.DataFrame({'ID' : ['O60829','O60341','Q9H1R3'],'TOTAL_COVERAGE' : ['yes','yes','no'],'BEG_D' : ['1','1','500'],'END_D' : ['102','25','600'],'BEG_S' : ['1','1'],'END_S': ['102','458']})
df = (df
.assign(pct=lambda x: np.where(x["TOTAL_COVERAGE"].eq("yes"),(x['END_S'].astype(int)*100)/x['END_D'].astype(int),np.nan))
.rename(columns={"pct":"%"})
)
输出
ID TOTAL_COVERAGE BEG_D END_D BEG_S END_S %
O60829 yes 1 102 1 102 100.0
O60341 yes 1 25 1 25 100.0
Q9H1R3 no 500 600 1 458 NaN