通过列表理解将str.lower应用于熊猫

Question

I have a dataframe df of the form:

    animal    fruit
0    "Dog"    "Apple"
1    "Cat"    "Banana"
2    "Rat"    "Grape"

I want to apply str.lower() to all columns (but not headers).

This Works:

for i in df:
    df[i] = df[i].str.lower()

How can I write this as a list comphrension?

I tried:

df[i] = [df[i].str.lower() for i in df]

But this does not work and I get a:

TypeError: list indices must be integers, not instancemethod

What must I change within the list comprehension for this to work?

Secondly, is there a more "Pandas-onicy" way of doing this in general, perhaps using the pandas.apply() function?

Many thanks for your help.

Answer 1

Output from list comprehension is list of Series . So need concat list :

L = [df[i].str.lower() for i in df]
print (L)
[0    dog
1    cat
2    rat
Name: animal, dtype: object, 0     apple
1    banana
2     grape
Name: fruit, dtype: object]

df1 = pd.concat(L, axis=1)
print (df1)
  animal   fruit
0    dog   apple
1    cat  banana
2    rat   grape

---

Solution with apply :

print (df.apply(lambda x: x.str.lower()))
  animal   fruit
0    dog   apple
1    cat  banana
2    rat   grape

Timings :

df = pd.concat([df]*1000).reset_index(drop=True)
df = pd.concat([df]*1000, axis=1)
df.columns = range(len(df.columns))
#[3000 rows x 2000 columns]
print (df)

In [89]: %timeit (pd.concat([df[i].str.lower() for i in df], axis=1))
1 loop, best of 3: 2.3 s per loop

In [90]: %timeit (df.apply(lambda x: x.str.lower()))
1 loop, best of 3: 2.63 s per loop

In [91]: %timeit (df.stack().str.lower().unstack())
1 loop, best of 3: 5.04 s per loop

Answer 2

You can stack so that it makes a single column, then call str.lower , and then unstack to restore the columns back:

In [74]:
df = df.stack().str.lower().unstack()
df

Out[74]:
  animal   fruit
0    dog   apple
1    cat  banana
2    rat   grape