pandas np.where 多个返回值
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np.where multiple return values
提问by DGraham
Using pandas and numpy I am trying to process a column in a dataframe, and want to create a new column with values relating to it. So if in column x the value 1 is present, in the new column it would be a, for value 2 it would be b etc
使用 pandas 和 numpy 我正在尝试处理数据框中的一列，并希望创建一个包含与其相关的值的新列。因此，如果在 x 列中存在值 1，则在新列中它将是 a，对于值 2 它将是 b 等
I can do this for single conditions, i.e
我可以针对单个条件执行此操作，即
df['new_col'] = np.where(df['col_1'] == 1, a, n/a)
And I can find example of multiple conditions i.e if x = 3 or x = 4 the value should a, but not to do something like if x = 3 the value should be a and if x = 4 the value be c.
我可以找到多个条件的示例，即如果 x = 3 或 x = 4，则该值应该是 a，但不要做类似如果 x = 3 的值应该是 a 并且如果 x = 4 的值是 c 的事情。
I tried simply running two lines of code such as :
我尝试简单地运行两行代码，例如：
df['new_col'] = np.where(df['col_1'] == 1, a, n/a)
df['new_col'] = np.where(df['col_1'] == 2, b, n/a)
But obviously the second line overwrites. Am I missing something crucial?
但显然第二行会覆盖。我错过了一些重要的东西吗？
回答by jezrael
回答by Stop harming Monica
I think numpy choose()is the best option for you.
我认为 numpychoose()是您的最佳选择。
import numpy as np
choices = 'abcde'
N = 10
np.random.seed(0)
data = np.random.randint(1, len(choices) + 1, size=N)
print(data)
print(np.choose(data - 1, choices))
Output:
输出：
[5 1 4 4 4 2 4 3 5 1]
['e' 'a' 'd' 'd' 'd' 'b' 'd' 'c' 'e' 'a']
回答by SpeedCoder5
Use the pandas Series.mapinstead of where.
使用 pandas Series.map而不是 where。
import pandas as pd
df = pd.DataFrame({'col_1' : [1,2,4,2]})
print(df)
def ab_ify(v):
if v == 1:
return 'a'
elif v == 2:
return 'b'
else:
return None
df['new_col'] = df['col_1'].map(ab_ify)
print(df)
# output:
#
# col_1
# 0 1
# 1 2
# 2 4
# 3 2
# col_1 new_col
# 0 1 a
# 1 2 b
# 2 4 None
# 3 2 b
回答by rde
you could define a dict with your desired transformations. Then loop through the a DataFrame column and fill it.
你可以用你想要的转换定义一个字典。然后循环遍历 DataFrame 列并填充它。
There may a more elegant ways, but this will work:
可能有更优雅的方法，但这会起作用：
# create a dummy DataFrame
df = pd.DataFrame( np.random.randint(2, size=(6,4)), columns=['col_1', 'col_2', 'col_3', 'col_4'], index=range(6) )
# create a dict with your desired substitutions:
swap_dict = { 0 : 'a',
1 : 'b',
999 : 'zzz', }
# introduce new column and fill with swapped information:
for i in df.index:
df.loc[i, 'new_col'] = swap_dict[ df.loc[i, 'col_1'] ]
print df
returns something like:
返回类似：
col_1 col_2 col_3 col_4 new_col
0 1 1 1 1 b
1 1 1 1 1 b
2 0 1 1 0 a
3 0 1 0 0 a
4 0 0 1 1 a
5 0 0 1 0 a
