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BUG: LatexFormatter.write_result multi-index #17499
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@@ -415,6 +415,7 @@ I/O | |
- Bug in :func:`read_stata` where value labels could not be read when using an iterator (:issue:`16923`) | ||
- Bug in :func:`read_html` where import check fails when run in multiple threads (:issue:`16928`) | ||
- Bug in :func:`read_csv` where automatic delimiter detection caused a ``TypeError`` to be thrown when a bad line was encountered rather than the correct error message (:issue:`13374`) | ||
- Bug in :func:`to_latex` where repeated multi-index values were not printed even though a higher level index differed from the previous row (:issue:`14484`) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. can you move to 0.21.1 |
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Plotting | ||
^^^^^^^^ | ||
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Original file line number | Diff line number | Diff line change |
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@@ -45,7 +45,6 @@ | |
import pandas as pd | ||
import numpy as np | ||
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import itertools | ||
import csv | ||
from functools import partial | ||
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@@ -887,6 +886,7 @@ def get_col_type(dtype): | |
name = any(self.frame.index.names) | ||
cname = any(self.frame.columns.names) | ||
lastcol = self.frame.index.nlevels - 1 | ||
previous_lev3 = None | ||
for i, lev in enumerate(self.frame.index.levels): | ||
lev2 = lev.format() | ||
blank = ' ' * len(lev2[0]) | ||
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@@ -897,11 +897,19 @@ def get_col_type(dtype): | |
lev3 = [blank] * clevels | ||
if name: | ||
lev3.append(lev.name) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I think it might be better to restructure this whole loop. iterating like this:
would be simpler to print out I think. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. You are correct, using df.groupby makes it simpler. Change is implemented and I will ping you once it is green. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I just had to revert my change. My attempt using df.groupby() changed the order of some elements and it didn't format more complex indicies like dates correctly. The current approach does not have any of these issues, so I think it is the best I can do right now. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ok, neither of these should be true, come back when you can There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ok thanks There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. |
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for level_idx, group in itertools.groupby( | ||
self.frame.index.labels[i]): | ||
count = len(list(group)) | ||
lev3.extend([lev2[level_idx]] + [blank] * (count - 1)) | ||
current_idx_val = None | ||
for level_idx in self.frame.index.labels[i]: | ||
if ((previous_lev3 is None or | ||
previous_lev3[len(lev3)].isspace()) and | ||
lev2[level_idx] == current_idx_val): | ||
# same index as above row and left index was the same | ||
lev3.append(blank) | ||
else: | ||
# different value than above or left index different | ||
lev3.append(lev2[level_idx]) | ||
current_idx_val = lev2[level_idx] | ||
strcols.insert(i, lev3) | ||
previous_lev3 = lev3 | ||
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column_format = self.column_format | ||
if column_format is None: | ||
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@@ -217,6 +217,28 @@ def test_to_latex_multiindex(self): | |
1 & 2.0 & 3.5 & 0.707107 & 3.0 & 3.25 & 3.5 & 3.75 & 4.0 \\ | ||
\bottomrule | ||
\end{tabular} | ||
""" | ||
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assert result == expected | ||
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# GH 14484 | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Add more comments about what you're testing here. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. can you make this a separate test |
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# If an index is repeated in subsequent rows, it should be | ||
# replaced with a blank in the created table. | ||
# This should ONLY happen if all higher order indicies | ||
# (to the left) are equal, too. | ||
# In this test, 'c' has to be printed both times, because | ||
# the higher order index 'A' != 'B' | ||
df = pd.DataFrame(index=pd.MultiIndex.from_tuples( | ||
[('A', 'c'), ('B', 'c')]), columns=['col']) | ||
result = df.to_latex() | ||
expected = r"""\begin{tabular}{lll} | ||
\toprule | ||
& & col \\ | ||
\midrule | ||
A & c & NaN \\ | ||
B & c & NaN \\ | ||
\bottomrule | ||
\end{tabular} | ||
""" | ||
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assert result == expected | ||
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I think we might have to simplify this, but I'll see what @jreback thinks.