问题
I have a pandas dataframe and I need to create a new column based on an if-else condition. This question already came up here multiple times (e.g., Creating a new column based on if-elif-else condition).
However, I cannot apply the proposed solution, since I also need to look up values in a list in order to check the condition. I cannot do this with the proposed solution, because I am not sure how I can access my lookup-list in the external function. My lookup-list would need to be global, which I want to avoid. I have the feeling there should be a better way to do this.
Consider the following dataframe df:
letters
A
B
C
D
E
F
I also have a list which contains lookup values:
lookup = [C,D]
Now, I want to create a new column in my dataframe which contains 1 if the respective value is contained in lookup and 0 if the values is not in lookup.
The typical approach would be:
df.apply(helper, axis=1)
def helper(row):
if(row['letters'].isin(lookup)):
row['result'] = 1
else:
row['result'] = 0
However, I do not know how I can access lookup in helper() without making it global.
The result should look like this:
letters result
A 0
B 0
C 1
D 1
E 0
F 0
回答1:
Although this question is very similar to the question: How to use pandas apply function on all columns of some rows of data frame
I think here it's worth showing a couple methods, on a single line using np.where with a boolean mask generated from isin, isin will return a boolean Series where any rows contain any matches in your list:
In [71]:
lookup = ['C','D']
df['result'] = np.where(df['letters'].isin(lookup), 1, 0)
df
Out[71]:
letters result
0 A 0
1 B 0
2 C 1
3 D 1
4 E 0
5 F 0
here using 2 loc statements and using ~ to invert the mask:
In [72]:
df.loc[df['letters'].isin(lookup),'result'] = 1
df.loc[~df['letters'].isin(lookup),'result'] = 0
df
Out[72]:
letters result
0 A 0
1 B 0
2 C 1
3 D 1
4 E 0
5 F 0
来源：https://stackoverflow.com/questions/38501685/create-new-pandas-dataframe-column-based-on-if-else-condition-with-a-lookup