BUG: decimal=',' does not work properly with complex values

[pjrandewijk]

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

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

• (optional) I have confirmed this bug exists on the master branch of pandas.

---

Note: Please read this guide detailing how to provide the necessary information for us to reproduce your bug.

Code Sample, a copy-pastable example

# Your code here
import pandas as pd
Y0=array([[0.-20.j        , 0. +0.j ],
                [0. +0.j        , 0.-12.j  ]])
Y0df=pd.DataFrame(Y0)
print(Y0df.to_latex(header=False,index=False,decimal=','), file=open('Y0.tex', 'w'))

Problem description

The contence of Y0.tex is shown below with only the real part having the correct decimal symbol.

\begin{tabular}{rr}
\toprule
0,000000-20.000000j & 0,000000+0.000000j \
0,000000+0.000000j & 0,000000-12.000000j \
\bottomrule
\end{tabular}

Expected Output

\begin{tabular}{rr}
\toprule
0,000000-20,000000j & 0,000000+0,000000j \
0,000000+0.000000j & 0,000000-12,000000j \
\bottomrule
\end{tabular}

Desired Output

#instead of using 'decimal' an 'use_numprint=True' would be nicer to allow for custom customisation within LaTeX, which would ideally produce the following output
%uncomment the following line if not already defined
%\newcommand{\np}[2][]{\numprint[#1]{#2}}
\begin{tabular}{rr}
\toprule
\np{0.000000}-\np{20.000000}j & \np{0.000000}+\np{0.000000}j \
\np{0.000000}+\np{0.000000}j & \np{0.000000}-\np{12.000000}j \
\bottomrule
\end{tabular}

Output of pd.show_versions()

INSTALLED VERSIONS

commit : db08276
python : 3.8.5.final.0
python-bits : 64
OS : Windows
OS-release : 10
Version : 10.0.18362
machine : AMD64
processor : Intel64 Family 6 Model 94 Stepping 3, GenuineIntel
byteorder : little
LC_ALL : None
LANG : en
LOCALE : English_South Africa.1252

pandas : 1.1.3
numpy : 1.19.2
pytz : 2020.1
dateutil : 2.8.1
pip : 20.2.4
setuptools : 50.3.1.post20201107
Cython : 0.29.21
pytest : 6.1.1
hypothesis : None
sphinx : 3.2.1
blosc : None
feather : None
xlsxwriter : 1.3.7
lxml.etree : 4.6.1
html5lib : 1.1
pymysql : None
psycopg2 : None
jinja2 : 2.11.2
IPython : 7.19.0
pandas_datareader: None
bs4 : 4.9.3
bottleneck : 1.3.2
fsspec : 0.8.3
fastparquet : None
gcsfs : None
matplotlib : 3.3.2
numexpr : 2.7.1
odfpy : None
openpyxl : 3.0.5
pandas_gbq : None
pyarrow : None
pytables : None
pyxlsb : None
s3fs : None
scipy : 1.5.2
sqlalchemy : 1.3.20
tables : 3.6.1
tabulate : None
xarray : None
xlrd : 1.2.0
xlwt : 1.3.0
numba : 0.51.2

[rhshadrach]

Thanks for the report! Could you open a new issue for the enhancement request?

Further investigations and PRs to fix are welcome!

[attack68]

If/When Styler.to_latex deprecates DataFrame.to_latex this fix is not required.
This will become available already after Styler.to_latex: #40422.
That PR specifically excludes formatting from the to_latex call for reasons stated in PR. The .format call is required.

Y0=np.array([[0.-20.j        , 0. +1000.j ],
             [0. +0.j        , 1000.-12.j  ]])
Y0df=pd.DataFrame(Y0)
print(Y0df.style.format(decimal=",", precision=2, thousands=".").to_latex())

\begin{tabular}{lrr}
{} & {0} & {1} \\
0 & 0,00-20,00j & 0,00+1.000,00j \\
1 & 0,00+0,00j & 1.000,00-12,00j \\
\end{tabular}

EDIT:
And here would be the solution to your requested format using a custom formatter:

def f(x):
    r = f"{x.real:,.3f}".replace(",", "#").replace(".", ",").replace("#", ".")
    i = f"{x.imag:,.3f}".replace(",", "#").replace(".", ",").replace("#", ".")
    return f"\\np{{{r}}}{'+' if x.imag >= 0 else '-'}\\np{{{i}}}j"

Y0=np.array([[0.-20.j        , 0. +1000.j ],
             [0. +0.j        , 1000.-12.j  ]])
Y0df=pd.DataFrame(Y0)
print(Y0df.style.format(f).to_latex())

\begin{tabular}{lrr}
{} & {0} & {1} \\
0 & \np{0,000}-\np{-20,000}j & \np{0,000}+\np{1.000,000}j \\
1 & \np{0,000}+\np{0,000}j & \np{1.000,000}-\np{-12,000}j \\
\end{tabular}