BUG: "styler.format.thousands" option doesn't work for integers

[vladpy8]

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
pandas.set_option('styler.format.thousands', ',')

data = pandas.DataFrame(data=[{'field': 1234567890}], dtype='int64',)

# Doesn't output integer field with comma separator
data

# Does output integer field with comma separator
data.style.format(thousands=',')

Issue Description

Printing data without .style in jupyter cell displays huge integer number without comma separator, inspite of the option set
Printing data with .style in jupyter cell displays integers with commas

Expected Behavior

Outputs must be same

Installed Versions

INSTALLED VERSIONS

commit : d9cdd2e
python : 3.11.9.final.0
python-bits : 64
OS : Darwin
OS-release : 23.4.0
Version : Darwin Kernel Version 23.4.0: Fri Mar 15 00:10:42 PDT 2024; root:xnu-10063.101.17~1/RELEASE_ARM64_T6000
machine : arm64
processor : arm
byteorder : little
LC_ALL : None
LANG : None
LOCALE : None.UTF-8

pandas : 2.2.2
numpy : 1.26.4
pytz : 2024.1
dateutil : 2.9.0.post0
setuptools : 69.5.1
pip : 24.0
Cython : None
pytest : None
hypothesis : None
sphinx : None
blosc : None
feather : None
xlsxwriter : None
lxml.etree : None
html5lib : None
pymysql : None
psycopg2 : None
jinja2 : 3.1.4
IPython : 8.24.0
pandas_datareader : None
adbc-driver-postgresql: None
adbc-driver-sqlite : None
bs4 : None
bottleneck : None
dataframe-api-compat : None
fastparquet : None
fsspec : 2024.5.0
gcsfs : None
matplotlib : 3.9.0
numba : None
numexpr : None
odfpy : None
openpyxl : None
pandas_gbq : None
pyarrow : 10.0.1
pyreadstat : None
python-calamine : None
pyxlsb : None
s3fs : 2024.5.0
scipy : 1.13.0
sqlalchemy : None
tables : None
tabulate : 0.9.0
xarray : None
xlrd : None
zstandard : None
tzdata : 2024.1
qtpy : None
pyqt5 : None

[attack68]

Agreed. This needs fixing, also for decimal option.

The solution, I believe, is in style_render:

#edited out

change to:

#edited out

and replace all the defaultdicts with the new ddict.

Needs tests.

[attack68]

The above was not correct. The following works:

pd.set_option('styler.format.precision', 3)
pd.set_option('styler.format.thousands', "+")
pd.set_option('styler.format.decimal', "-")
df = pd.DataFrame([
    [10000.0, 20000],
    [30000.0, 40000]
])
df.style

as does

pd.set_option('styler.format.precision', 3)
pd.set_option('styler.format.thousands', "+")
pd.set_option('styler.format.decimal', "-") 
df = pd.DataFrame(data=[{'field': 1234567890}], dtype='int64',)
df.style

I misread your issue. The styler pandas options only impact the Styler class. These options are available for people who regularly use Styler. They do not impact DataFrame and the options associated with that class and any of its data input, wrangling, or (in your case) output methods. This is desired behaviour.
If your do not call .style you will not produce a Styler and these options will not apply.

[mroeschke]

Yeah given the above it seems like the OP is the expected behavior so closing

[vladpy8]

The styler pandas options only impact the Styler class.

They do not impact DataFrame and the options associated with that class and any of its data input, wrangling, or (in your case) output methods.

I see, thank you!