当存在多个变量时，使用lambda在python数据帧中实现if-else

Question

I am trying to implement if-elif or if-else logic in python while working on a dataframe. I am struggling when working with more than one column.

sample data frame

df=pd.DataFrame({"one":[1,2,3,4,5],"two":[6,7,8,9,10], "name": 'a', 'b', 'a', 'b', 'c'})

If my if-else logic is based on only one column - I know how to do it.

df['one'] = df["one"].apply(lambda x: x*10 if x<2 else (x**2 if x<4 else x+10))

But I want to modify column 'one' based on values of column 'two' - and I feel its going be something like this -

lambda x, y: x*100 if y>8 else (x*1 if y<8 else x**2)

But I am not sure how to specify the second column. I tried this way but obviously that's incorrect

df['one'] = df["one"]["two"].apply(lambda x, y: x*100 if y>8 else (x*1 if y<8 else x**2))

Question 1 - what'd be the correct syntax for the above code ?

Question 2 - How to implement below logic using lambda ?

if df['name'].isin(['a','b'])  df['one'] = 100 else df['one'] = df['two']

If I write something like x.isin(['a','b']) it won't work.

Answer 1

Apply across columns

Use pd.DataFrame.apply instead of pd.Series.apply and specify axis=1 :

df['one'] = df.apply(lambda row: row['one']*100 if row['two']>8 else \
                     (row['one']*1 if row['two']<8 else row['one']**2), axis=1)

Unreadable? Yes, I agree. Let's try again but this time rewrite as a named function.

Using a function

Note lambda is just an anonymous function. We can define a function explicitly and use it with pd.DataFrame.apply :

def calc(row):
    if row['two'] > 8:
        return row['one'] * 100
    elif row['two'] < 8:
        return row['one']
    else:
        return row['one']**2

df['one'] = df.apply(calc, axis=1)

Readable? Yes. But this isn't vectorised. We're looping through each row one at at at time. We might as well have used a list. Pandas isn't just for clever table formatting, you can use it for vectorised calculations using arrays in contiguous memory blocks. So let's try one more time.

Vectorised calculations

Using numpy.where :

df['one'] = np.where(row['two'] > 8, row['one'] * 100,
                     np.where(row['two'] < 8, row['one'],
                              row['one']**2))

There we go. Readable and efficient. We have effectively vectorised our if / else statements. Does this mean that we are doing more calculations than necessary? Yes! But this is more than offset by the way in which we are performing the calculations, ie with well-defined blocks of memory rather than pointers. You will find an order of magnitude performance improvement.

Another example

Well, we can just use numpy.where again.

df['one'] = np.where(df['name'].isin(['a', 'b']), 100, df['two'])

Answer 2

you can do

df.apply(lambda x: x["one"] + x["two"], axis=1)

but i don't think that such a long lambda as lambda x: x["one"]*100 if x["two"]>8 else (x["one"]*1 if x["two"]<8 else x["one"]**2) is very pythonic. apply takes any callback:

def my_callback(x):
    if x["two"] > 8:
        return x["one"]*100
    elif x["two"] < 8:
        return x["one"]
    else:
        return x["one"]**2

df.apply(my_callback, axis=1)