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BUG: to_latex() does not handle braces in new headers gracefully #60063

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@GregersSR

Description

@GregersSR

Pandas version checks

  • I have checked that this issue has not already been reported.

  • I have confirmed this bug exists on the latest version of pandas.

  • I have confirmed this bug exists on the main branch of pandas.

Reproducible Example

import pandas as pd

df = pd.DataFrame({"a": [1, 2]})
result = df.to_latex(header=[r"$\bar{y}$"])

Issue Description

This gives a KeyError. I looked into the reason, and it happens because at some point r"$\bar{y}$".format(x) is called, and the braces are interpreted as string interpolation.

Expected Behavior

I expect one of the following:

  1. An error message saying that single braces should be escaped using {{
  2. Documentation stating the same thing
  3. That the single brace is not interpreted as string interpolation

Installed Versions

pd.show_versions()

INSTALLED VERSIONS

commit : 2a10e04
python : 3.10.15
python-bits : 64
OS : Linux
OS-release : 6.8.0-45-generic
Version : #45-Ubuntu SMP PREEMPT_DYNAMIC Fri Aug 30 12:02:04 UTC 2024
machine : x86_64
processor : x86_64
byteorder : little
LC_ALL : None
LANG : en_US.UTF-8
LOCALE : en_US.UTF-8

pandas : 3.0.0.dev0+1579.g2a10e04a09
numpy : 1.26.4
dateutil : 2.9.0
pip : 24.2
Cython : 3.0.11
sphinx : 8.1.3
IPython : 8.28.0
adbc-driver-postgresql: None
adbc-driver-sqlite : None
bs4 : 4.12.3
blosc : None
bottleneck : 1.4.1
fastparquet : 2024.5.0
fsspec : 2024.9.0
html5lib : 1.1
hypothesis : 6.115.2
gcsfs : 2024.9.0post1
jinja2 : 3.1.4
lxml.etree : 5.3.0
matplotlib : 3.9.2
numba : 0.60.0
numexpr : 2.10.1
odfpy : None
openpyxl : 3.1.5
psycopg2 : 2.9.9
pymysql : 1.4.6
pyarrow : 17.0.0
pyreadstat : 1.2.7
pytest : 8.3.3
python-calamine : None
pytz : 2024.2
pyxlsb : 1.0.10
s3fs : 2024.9.0
scipy : 1.14.1
sqlalchemy : 2.0.36
tables : 3.10.1
tabulate : 0.9.0
xarray : 2024.9.0
xlrd : 2.0.1
xlsxwriter : 3.2.0
zstandard : 0.23.0
tzdata : 2024.2
qtpy : None
pyqt5 : None

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BugIO LaTeXto_latexStylerconditional formatting using DataFrame.style

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