从 Pandas DataFrame 列标题中获取列表

Question

I want to get a list of the column headers from a Pandas DataFrame. The DataFrame will come from user input, so I won't know how many columns there will be or what they will be called.

For example, if I'm given a DataFrame like this:

>>> my_dataframe
    y  gdp  cap
0   1    2    5
1   2    3    9
2   8    7    2
3   3    4    7
4   6    7    7
5   4    8    3
6   8    2    8
7   9    9   10
8   6    6    4
9  10   10    7

I would get a list like this:

>>> header_list
['y', 'gdp', 'cap']

Answer 1

You can get the values as a list by doing:

list(my_dataframe.columns.values)

Also you can simply use: (as shown in Ed Chum's answer ):

list(my_dataframe)

Answer 2

There is a built in method which is the most performant:

my_dataframe.columns.values.tolist()

.columns returns an Index, .columns.values returns an array and this has a helper function .tolist to return a list.

If performance is not as important to you, Index objects define a .tolist() method that you can call directly:

my_dataframe.columns.tolist()

The difference in performance is obvious:

%timeit df.columns.tolist()
16.7 µs ± 317 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit df.columns.values.tolist()
1.24 µs ± 12.3 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

---

For those who hate typing, you can just call list on df , as so:

list(df)

Answer 3

Did some quick tests, and perhaps unsurprisingly the built-in version using dataframe.columns.values.tolist() is the fastest:

In [1]: %timeit [column for column in df]
1000 loops, best of 3: 81.6 µs per loop

In [2]: %timeit df.columns.values.tolist()
10000 loops, best of 3: 16.1 µs per loop

In [3]: %timeit list(df)
10000 loops, best of 3: 44.9 µs per loop

In [4]: % timeit list(df.columns.values)
10000 loops, best of 3: 38.4 µs per loop

(I still really like the list(dataframe) though, so thanks EdChum!)

Answer 4

Its gets even simpler (by pandas 0.16.0) :

df.columns.tolist()

will give you the column names in a nice list.

Answer 5

Surprised I haven't seen this posted so far, so I'll just leave this here.

Extended Iterable Unpacking (python3.5+): [*df] and Friends

Unpacking generalizations (PEP 448) have been introduced with Python 3.5. So, the following operations are all possible.

df = pd.DataFrame('x', columns=['A', 'B', 'C'], index=range(5))
df

   A  B  C
0  x  x  x
1  x  x  x
2  x  x  x
3  x  x  x
4  x  x  x 

If you want a list ....

[*df]
# ['A', 'B', 'C']

Or, if you want a set ,

{*df}
# {'A', 'B', 'C'}

Or, if you want a tuple ,

*df,  # Please note the trailing comma
# ('A', 'B', 'C')

Or, if you want to store the result somewhere,

*cols, = df  # A wild comma appears, again
cols
# ['A', 'B', 'C']

... if you're the kind of person who converts coffee to typing sounds, well, this is going consume your coffee more efficiently ;)

PS: if performance is important, you will want to ditch the solutions above in favour of

df.columns.to_numpy().tolist() # ['A', 'B', 'C']

This is similar to Ed Chum's answer , but updated for v0.24 where .to_numpy() is preferred to the use of .values . See this answer (by me) for more information.

Visual Check
Since I've seen this discussed in other answers, you can utilise iterable unpacking (no need for explicit loops).

print(*df)
A B C

print(*df, sep='\n')
A
B
C

---

Critique of Other Methods

Don't use an explicit for loop for an operation that can be done in a single line (List comprehensions are okay).

Next, using sorted(df) does not preserve the original order of the columns. For that, you should use list(df) instead.

Next, list(df.columns) and list(df.columns.values) are poor suggestions (as of the current version, v0.24). Both Index (returned from df.columns ) and NumPy arrays (returned by df.columns.values ) define .tolist() method which is faster and more idiomatic.

Lastly, listification ie, list(df) should only be used as a concise alternative to the aforementioned methods for python <= 3.4 where extended unpacking is not available.

Answer 6

>>> list(my_dataframe)
['y', 'gdp', 'cap']

To list the columns of a dataframe while in debugger mode, use a list comprehension:

>>> [c for c in my_dataframe]
['y', 'gdp', 'cap']

By the way, you can get a sorted list simply by using sorted :

>>> sorted(my_dataframe)
['cap', 'gdp', 'y']

Answer 7

这可以作为my_dataframe.columns 。

Answer 8

ADataFrame follows the dict-like convention of iterating over the “keys” of the objects.

my_dataframe.keys()

Create a list of keys/columns - object method to_list() and pythonic way

my_dataframe.keys().to_list()
list(my_dataframe.keys())

Basic iteration on a DataFrame returns column labels

[column for column in my_dataframe]

Do not convert a DataFrame into a list, just to get the column labels. Do not stop thinking while looking for convenient code samples.

xlarge = pd.DataFrame(np.arange(100000000).reshape(10000,10000))
list(xlarge) #compute time and memory consumption depend on dataframe size - O(N)
list(xlarge.keys()) #constant time operation - O(1)

Answer 9

It's interesting but df.columns.values.tolist() is almost 3 times faster then df.columns.tolist() but I thought that they are the same:

In [97]: %timeit df.columns.values.tolist()
100000 loops, best of 3: 2.97 µs per loop

In [98]: %timeit df.columns.tolist()
10000 loops, best of 3: 9.67 µs per loop

Answer 10

In the Notebook

For data exploration in the IPython notebook, my preferred way is this:

sorted(df)

Which will produce an easy to read alphabetically ordered list.

In a code repository

In code I find it more explicit to do

df.columns

Because it tells others reading your code what you are doing.

Answer 11

%%timeit
final_df.columns.values.tolist()
948 ns ± 19.2 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%%timeit
list(final_df.columns)
14.2 µs ± 79.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
list(final_df.columns.values)
1.88 µs ± 11.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%%timeit
final_df.columns.tolist()
12.3 µs ± 27.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%%timeit
list(final_df.head(1).columns)
163 µs ± 20.6 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 12

I want to get a list of the column headers from a pandas DataFrame. The DataFrame will come from user input so I won't know how many columns there will be or what they will be called.

For example, if I'm given a DataFrame like this:

>>> my_dataframe
    y  gdp  cap
0   1    2    5
1   2    3    9
2   8    7    2
3   3    4    7
4   6    7    7
5   4    8    3
6   8    2    8
7   9    9   10
8   6    6    4
9  10   10    7

I would get a list like this:

>>> header_list
['y', 'gdp', 'cap']

Answer 13

as answered by Simeon Visser...you could do

list(my_dataframe.columns.values) 

or

list(my_dataframe) # for less typing.

But I think most the sweet spot is:

list(my_dataframe.columns)

It is explicit, at the same time not unnecessarily long.

Answer 14

For a quick, neat, visual check, try this:

for col in df.columns:
    print col

Answer 15

I feel question deserves additional explanation.

As @fixxxer noted, the answer depends on the pandas version you are using in your project. Which you can get with pd.__version__ command.

If you are for some reason like me (on debian jessie I use 0.14.1) using older version of pandas than 0.16.0, then you need to use:

df.keys().tolist() because there is no df.columns method implemented yet.

The advantage of this keys method is, that it works even in newer version of pandas, so it's more universal.

Answer 16

import pandas as pd

# create test dataframe
df = pd.DataFrame('x', columns=['A', 'B', 'C'], index=range(2))

list(df.columns)

Returns

['A', 'B', 'C']

Answer 17

n = []
for i in my_dataframe.columns:
    n.append(i)
print n

Answer 18

I want to get a list of the column headers from a pandas DataFrame. The DataFrame will come from user input so I won't know how many columns there will be or what they will be called.

For example, if I'm given a DataFrame like this:

>>> my_dataframe
    y  gdp  cap
0   1    2    5
1   2    3    9
2   8    7    2
3   3    4    7
4   6    7    7
5   4    8    3
6   8    2    8
7   9    9   10
8   6    6    4
9  10   10    7

I would get a list like this:

>>> header_list
['y', 'gdp', 'cap']

Answer 19

Even though the solution that was provided above is nice. I would also expect something like frame.column_names() to be a function in pandas, but since it is not, maybe it would be nice to use the following syntax. It somehow preserves the feeling that you are using pandas in a proper way by calling the "tolist" function: frame.columns.tolist()

frame.columns.tolist() 

Answer 20

If the DataFrame happens to have an Index or MultiIndex and you want those included as column names too:

names = list(filter(None, df.index.names + df.columns.values.tolist()))

It avoids calling reset_index() which has an unnecessary performance hit for such a simple operation.

I've run into needing this more often because I'm shuttling data from databases where the dataframe index maps to a primary/unique key, but is really just another "column" to me. It would probably make sense for pandas to have a built-in method for something like this (totally possible I've missed it).

Answer 21

listHeaders = [my_dataframe 中 colName 的 colName]

Answer 22

The simplest option would be: list(my_dataframe.columns) or my_dataframe.columns.tolist()

No need for the complex stuff above :)

Answer 23

This is the easiest way to reach your goal.

my_dataframe.columns.values.tolist()

and if you are Lazy, try this >

list(my_dataframe)

Answer 24

its the simple code for you:

for i in my_dataframe:
    print(i)

just do it

Answer 25

Its very simple.

Like you can do it as:

list(df.columns)

Answer 26

此解决方案列出了对象 my_dataframe 的所有列：

print(list(my_dataframe))