'How to justify text in label in Tkinter

In Tkinter in Python: I have a table with a different label. How can I justify the text that is in the label? Because It is a table and the texts in different labels come together!

from tkinter import *
root=Tk()
a=Label(root,text='Hello World!')
a.pack()
a.place(x=200,y=200)
b=Label(root,text='Bye World')
b.pack()
b.place(x=200,y=100)

I want something for justifying in center some text in label but it is not something that I need plz check this: link



Solution 1:[1]

By default, the text in a label is centered in the label. You can control this with the anchor attribute, and possibly with the justify attribute. justify only affects the text when there is more than one line of text in the widget.

For example, to get the text inside a label to be right-aligned you can use anchor="e":

a=Label(root,text='Hello World!', anchor="e")

Note, however, this will appear to have no effect if the label is exactly big enough to hold the text. In your specific example, you would need to give each label the same width:

a=Label(..., width=12)
b=Label(..., width=12)

Solution 2:[2]

To add on to what Bryan said, LEFT is the constant you are looking for to correctly format your wrapped text. You can also justify to the RIGHT or CENTER (the default).

a=Label(root,text='Hello World!', anchor="e", justify=LEFT)

Solution 3:[3]

instead of using .pack() i would use .grid()

The Tkinter Grid Geometry Manager

grid will allow better management of your components

find bellow an example of usage and management:

Label(root, text="First").grid(row=0, sticky=W)
Label(root, text="Second").grid(row=1, sticky=W)

entry1 = Entry(root)
entry2 = Entry(root)

entry1.grid(row=0, column=1)
entry2.grid(row=1, column=1)

checkbutton.grid(columnspan=2, sticky=W)

image.grid(row=0, column=2, columnspan=2, rowspan=2,
           sticky=W+E+N+S, padx=5, pady=5)

button1.grid(row=2, column=2)
button2.grid(row=2, column=3)

you would endup using the grid option padx="x" to "justify" your labels

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Bryan Oakley
Solution 2 Vincent Wetzel
Solution 3 Alex Peters