[英]Multi Conditional if Statement Pandas (code doesn't finish running)
尝试根据其他列值在 pandas 中创建多条件新列。
以下代码不会产生错误或结果(但会产生警告); 它一直在运行:
for val1,val2 in zip(df['a'], df['b']):
if val1 == 0 and val2 == 0:
df['new_column'] = 0
elif val1 in df['a'] == 0:
df['new_column'] = 1
else:
for val2 in df['b']:
if val2 ==0:
df['new_column'] = 0
else:
df['new_column'] = df['b'] / df['a']
df 看起来像:
['a'] ['b']
0 0
0 1000
1000 0
5000 2000
期望 df['new column'] 是这样的:
['new column']
0
1
0
.4
这可能不是最优雅的解决方案,但根据问题中提供的有限信息,此代码至少会生成您预期的输出:
def myfunc(row):
if row['a'] == 0 and row['b'] == 0:
result = 0
else:
if row['a'] == 0:
result = 1
elif row['b'] == 0:
result = 0
else:
result = row['b'] / row['a']
return result
df['new column'] = df.apply(myfunc, axis=1)
不要使用循环/ apply
,使用矢量代码:
df['new'] = (df['b'].div(df['a'].mask(df['a'].eq(0),df['b']))
.fillna(0)
)
输出:
a b new
0 0 0 0.0
1 0 1000 1.0
2 1000 0 0.0
3 5000 2000 0.4
与@cucurbit 在细节上有所不同,但原理相同:
def vals(x, y):
if x == 0 and y == 0:
return 0
if x == 0 :
return 1
if y == 0:
return 0
else:
return y/x
df['new column'] = df.apply(lambda x : vals(x['a'], x['b']), axis =1)
事实证明,我所从事的工作特别需要的是利用嵌套的 np.where 语句。 定义一个函数并应用它并没有为我产生正确的计算结果。
df['new_column'] = np.where((df['a'] == 0) & (df['b'] == 0), 0,
np.where(df['a'] == 0, 1,
np.where(df['b'] ==-0, 0, df['b']/df['a']
)))
我感谢大家的意见。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.