How to display a specific column's headers name using the panda package in Python

Question

I want to display just the column's header name. I tried using .columns, but that list all the data frame's column headers and I just want to only print the header of the two columns I'm passing.

def test(x, y): # x and y are specific columns within a data frame that is being passed.

   print(x.columns, y.columns) # Runs an error of course, but I want to know how I would display just display their header names.

So I used panda to create a data frame of a .csv, which could be this:

    num1   num2   num3
0    1      2      3
1    4      5      6
2    7      8      9

So lets say I passed the columns 'num1' and 'num3' as parameters to the function test as x and y. I want to print the names num1 and num3 since those are the header names. How would I go about that?

UPDATE: Okay so, I got the names to display using print(). However, I guess the actual error is when I do plt.xlabel(x.name) & plt.ylabel(y.name). It gives the error: 'str' object has no attribute 'pop'. I thought print() and plt.xlabel() or plt.ylabel() would be treated the same, my apologies.

UPDATE 2: Solved! Thanks guys!

an example would help people to understand what you want to achieve. — Dyno Fu
– Dyno Fu, Commented Oct 5, 2019 at 21:07
If you already know the names, why don't just print them? In your example you say you input 'num1' and 'num3' to print 'num1' and 'num3'. There is no look-up required you already have what you need. — Joe Ernst
– Joe Ernst, Commented Oct 5, 2019 at 21:37
@JoeErnst because I have a large list of columns/rows and I want to print multiple graphs. I would like to automate the labels for my graph without having to constantly change the names and just have it place it for me. — researchnewbie
– researchnewbie, Commented Oct 5, 2019 at 21:46

vb_rises · Accepted Answer · 2019-10-05 21:46:52Z

1

d = {'id': ['x1', 'x2'],'t1': [3,11]}
df = pd.DataFrame(data=d)


def test(x, y): # x and y are specific columns within a data frame that is being passed.
   print(x.name, y.name)
n1 = df['id']
n2 = df['t1']

test(n1,n2)

#output
id t1

edited Oct 5, 2019 at 21:46

answered Oct 5, 2019 at 21:34

vb_rises

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5 Comments

researchnewbie Over a year ago

Getting the error: 'str' object has no attribute 'pop'

vb_rises Over a year ago

@researchnewbie how you are calling the function?

researchnewbie Over a year ago

So what I did was I pretty much said n1 = df['num1'] and n2 = df['num2'], and then I called the function by doing test(n1, n2). My graph correctly displays the data, just not the names which I want to use as labels for plots.

vb_rises Over a year ago

@researchnewbie I have edited my answer. you can show more code.

researchnewbie Over a year ago

Okay, so how would I go about this? I don't really know what to modify from mines to get to this since I can't really recreate the data=d portion. Its over 100k rows and 75+ columns. The thing is with my data, I just entered the .csv and used read_csv, where it organized the data for me. It prints the graph correctly and identifies each header correctly as well, just won't do .name as it works in your code.

Massifox · Accepted Answer · 2019-10-05 21:55:40Z

1

Given the following dataframe:

df = pd.DataFrame([[1,2,3], [4,5,6], [7,8,9]], columns=['num1', 'num2', 'num3'])

Try this code:

def test(x, y):
    print(x.name, y.name)

test(x=df['num1'], y=df['num3'])

gives:

# output: num1 num3

Note: x and y are pandas series, and to get the name of the series (which corrsipsonde the name of the df column) you must use the attribute .name

If I understand your second question correctly, this code is for you:

cols = df.columns

for x, y in zip(cols, cols[1:]):
    test(df[x],df[y])

and gives:

num1 num2
num2 num3

edited Oct 5, 2019 at 21:55

answered Oct 5, 2019 at 21:35

Massifox

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3 Comments

researchnewbie Over a year ago

Do I have to define it like that when making it the arguments? I did it prior by doing this: x = df['num1] , y = df['num2'] and then test(x, y) yet it still doesn't work.

Massifox Over a year ago

@researchnewbie I updated the answer, let me know if that's what you mean

researchnewbie Over a year ago

I updated my answer to display more about whats going on.

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