如果一列中的值介于其他列中的两个值之间,则无法将权重(int)添加到新的Pandas DataFrame列。我能够创建具有True / False值的列(如果我使用astype,则为0/1值)。
import pandas as pd
df = pd.DataFrame({'a': [1,2,3], 'b': [4,5,6], 'c': [3,6,4]})
df
a b c
0 1 4 3
1 2 5 6
2 3 6 4
这有效:
df['between_bool'] = df['c'].between(df['a'], df['b'])
df
a b c between_bool
0 1 4 3 True # 3 is between 1 and 4
1 2 5 6 False # 6 is NOT between 2 and 5
2 3 6 4 True # 4 is between 3 and 6
然而,这不起作用:
df['between_int'] = df['c'].apply(lambda x: 2 if df['c'].between(df['a'], df['b']) else 0)
上面的代码会产生以下错误:
Traceback (most recent call last):
File "C:\Python36\envs\PortfolioManager\lib\site-packages\IPython\core\interactiveshell.py", line 2881, in run_code
exec(code_obj, self.user_global_ns, self.user_ns)
File "<ipython-input-14-0aa1e7cfd5c2>", line 1, in <module>
df['between_int'] = df['c'].apply(lambda x: 2 if df['c'].between(df['a'], df['b']) else 0)
File "C:\Python36\envs\PortfolioManager\lib\site-packages\pandas\core\series.py", line 2294, in apply
mapped = lib.map_infer(values, f, convert=convert_dtype)
File "pandas\src\inference.pyx", line 1207, in pandas.lib.map_infer (pandas\lib.c:66124)
File "<ipython-input-14-0aa1e7cfd5c2>", line 1, in <lambda>
所需的输出是:
a b c between_int
0 1 4 3 2 # 3 is between 1 and 4
1 2 5 6 0 # 6 is NOT between 2 and 5
2 3 6 4 2 # 4 is between 3 and 6
有什么想法吗?
答案 0 :(得分:1)
我希望我能正确理解你,但如果你只是想在这个条件下添加固定重量2,可以选择以下方法:
import numpy as np
df['between_int'] = np.where(df['c'].between(df['a'], df['b']), 2, 0)
或者,如果您不想导入numpy,则可以执行以下操作:
df['between_int'] = 0
df.loc[df['c'].between(df['a'], df['b']), 'between_int'] = 2
希望这有帮助!
答案 1 :(得分:1)
我认为您最初想要使用def func(data, (x_0,y_0)):
y, x = numpy.indices(data.shape)
r = (x - x_0)**2 + (y - y_0)**2
float_values, r = numpy.unique(r, return_inverse=True)
return float_values ** 0.5, r.reshape(data.shape)
做的是:
apply
看到与你的不同之处:
df['between_int'] = df.apply(lambda x: 2 if x['c'] in range(x['a'], x['b']) else 0, axis=1)
上的apply
而不是系列df
df['c']
而不是x['c']
检查的值,因为您的lambda是x的函数df['c']
更改为df['c']
我不能再使用x['c']
了between
in range
和x['a']
进行调用,原因与第2点相同x['b']
现在axis=1
在数据框无论如何,Swebbo的解决方案完美无缺!