Measuring string width in Python

Question

I have some font (Times New Roman, 14) and I need to measure string width. I found matplotlib and example from API:

from matplotlib import rcParams
import os.path
afm_filename = os.path.join(rcParams['datapath'], 'fonts', 'afm', 
               'ptmr8a.afm')
from matplotlib.afm import AFM
afm = AFM(open(afm_filename))
afm.string_width_height('What the heck?')

But I can't understand what I should paste instead of "datapaths", "fonts", "afm". Is it the way to get sting width in Python? Maybe another way?

My problem is to orginize some text in Word. So, I use python-docx for reading docx document. And I need to measure string width in cm in the document. If I use default settings (in code above): — Ivan
– Ivan, Commented Sep 10, 2017 at 11:06
Error: if not line.startswith(b'StartFontMetrics'): TypeError: startswith first arg must be str or a tuple of str, not bytes — Ivan
– Ivan, Commented Sep 10, 2017 at 11:07

RedEyed · Accepted Answer · 2017-09-10 12:24:04Z

3

import tkinter as Tkinter 
from tkinter import font as tkFont

Tkinter.Frame().destroy()
txt = tkFont.Font(family="Times New Roman", size=14)
width = txt.measure("What the heck?")
print(width)

edited Sep 10, 2017 at 12:24

answered Sep 10, 2017 at 11:00

RedEyed

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8 Comments

Ivan Over a year ago

Some error: File "E:/=ПРОЕКТЫ=/=PYTHON=/=ТЕКСТ=/Calc_String_Variants/Calc.py", line 4, in <module> txt = tkFont.Font(family="Times New Roman", size=14) File "C:\Users\User\AppData\Local\Programs\Python\Python36-32\lib\tkinter\font.py", line 93, in init tk.call("font", "create", self.name, *font) AttributeError: 'NoneType' object has no attribute 'call'

Ivan Over a year ago

it is some add:

RedEyed Over a year ago

Updated once more. Should work for python 2 and 3. What do you mean "it is some add" ?

Ivan Over a year ago

I add string "tkinter.Frame().destroy()" in Python 3 version and it is work, but it has problem: Width of my Word document is 505 mm. So if I add all letters "a" (230 letters) for full string - the answer is 1840, but if I add all letters "c" (230 letters) for full string - the answer is 2070. It is great difference, but in Word is about the same.

RedEyed Over a year ago

Updated with destroy()

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