如何使用Python Pandas在词典切片中选择数据的子集?
更多Pandas相关文章,请阅读:Pandas 教程
简介
Pandas具有双重选择能力,使用索引位置或使用索引标签选择数据子集。在本文中,我将向您展示如何“使用词典切片选择数据子集”。
谷歌上充满了数据集。在kaggle.com中搜索电影数据集。本文使用来自kaggle的电影数据集。
如何实现
- 导入仅用于此示例的列的电影数据集。
import pandas as pd
import numpy as np
movies = pd.read_csv("https://raw.githubusercontent.com/sasankac/TestDataSet/master/movies_data.csv",index_col="title",
usecols=["title","budget","vote_average","vote_count"])
movies.sample(n=5)
| budget | vote_average | vote_count | |
|---|---|---|---|
| titile | |||
| Little Voice | 0 | 6.6 | 61 |
| Grown Ups 2 | 80000000 | 5.8 | 1155 |
| The Best Years of Our Lives | 2100000 | 7.6 | 143 |
| Tusk | 2800000 | 5.1 | 366 |
| Operation Chromite | 0 | 5.8 | 29 |
- 我始终建议对索引进行排序,特别是如果索引由字符串组成。如果您的索引已排序,则在处理巨大数据集时会注意到差异。
如果我不对索引进行排序怎么办?
没问题,您的代码将永远运行。开个玩笑,如果索引标签未排序,则Pandas必须逐个遍历所有标签以匹配查询。想象一下没有索引页的牛津字典,你会怎么做?索引排序后,您可以快速跳转到要提取的标签,这也是Pandas的情况。
让我们首先检查索引是否已排序。
#检查索引是否已排序?
movies.index.is_monotonic
假
- 明显,索引未排序。我们将尝试选择以A%开头的电影。这就像写作
select * from movies where title like’A%’
movies.loc["Aa":"Bb"]
选择所有标题以'A%'开头的电影
---------------------------------------------------------------------------
ValueErrorTraceback (most recent call last)
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_slice_bound(self, label, side, kind)
4844try:
-> 4845return self._searchsorted_monotonic(label, side) 4846except ValueError:
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in _searchsorted_monotonic(self, label, side)
4805
-> 4806raise ValueError("index must be monotonic increasing or decreasing")
4807
ValueError: 索引必须单调递增或递减
在处理上述异常时,又发生了另一个异常:
KeyErrorTraceback (most recent call last)
in
----> 1 movies.loc["Aa": "Bb"]
~\anaconda3\lib\site-packages\pandas\core\indexing.py in getitem(self, key)
1766
1767maybe_callable = com.apply_if_callable(key, self.obj)
-> 1768return self._getitem_axis(maybe_callable, axis=axis) 1769
1770def _is_scalar_access(self, key: Tuple):
~\anaconda3\lib\site-packages\pandas\core\indexing.py in _getitem_axis(self, key, axis)
1910if isinstance(key, slice):
1911self._validate_key(key, axis)
-> 1912return self._get_slice_axis(key, axis=axis) 1913elif com.is_bool_indexer(key):
1914return self._getbool_axis(key, axis=axis)
~\anaconda3\lib\site-packages\pandas\core\indexing.py in _get_slice_axis(self, slice_obj, axis)
1794
1795labels = obj._get_axis(axis)
-> 1796indexer = labels.slice_indexer(
1797slice_obj.start, slice_obj.stop, slice_obj.step, kind=self.name 1798)
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in slice_indexer(self, start, end, step, kind)
4711slice(1, 3)
4712"""
-> 4713start_slice, end_slice = self.slice_locs(start, end, step=step, kind=kind)
4714
4715# return a slice
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in slice_locs(self, start, end, step, kind)
4924start_slice = None
4925if start is not None:
-> 4926start_slice = self.get_slice_bound(start, "left", kind) 4927if start_slice is None:
4928start_slice = 0
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_slice_bound(self, label, side, kind)
4846except ValueError:
4847# raise the original KeyError
-> 4848raise err
4849
4850if isinstance(slc, np.ndarray):
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_slice_bound(self, label, side, kind)
4840# we need to look up the label
4841try:
-> 4842slc = self.get_loc(label) 4843except KeyError as err:
4844try:
~\anaconda3\lib\site-packages\pandas\core\indexes\base.py in get_loc(self, key, method, tolerance)
2646return self._engine.get_loc(key)
2647except KeyError:
-> 2648return self._engine.get_loc(self._maybe_cast_indexer(key))
2649indexer = self.get_indexer([key], method=method, tolerance=tolerance)
2650if indexer.ndim > 1 or indexer.size > 1:
pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas\_libs\index.pyx in pandas._libs.index.IndexEngine.get_loc()
pandas\_libs\index.pyx in pandas._libs.index.IndexEngine._get_loc_duplicates()
pandas\_libs\index.pyx in pandas._libs.index.IndexEngine._maybe_get_bool_indexer()
KeyError: 'Aa'
- 将索引按升序排序,然后尝试使用词典排序切片的相同命令利用排序优势。
真
- 现在我们的数据设置好了,可以进行词典排序的切片。现在让我们选择所有从字母’A’到字母’B’开头的电影标题。
| budget | vote_average | vote_count | |
|---|---|---|---|
| title | |||
| Abandon | 25000000 | 4.6 | 45 |
| Abandoned | 0 | 5.8 | 27 |
| Abduction | 35000000 | 5.6 | 961 |
| Aberdeen | 0 | 7.0 | 6 |
| About Last Night | 12500000 | 6.0 | 210 |
| … | … | … | … |
| Battle for the Planet of the Apes | 1700000 | 5.5 | 215 |
| Battle of the Year | 20000000 | 5.9 | 88 |
| Battle: Los Angeles | 70000000 | 5.5 | 1448 |
| Battlefield Earth | 44000000 | 3.0 | 255 |
| Battleship | 209000000 | 5.5 | 2114 |
292 行 × 3列
True
| title | budget | vote_average | vote_count |
|---|---|---|---|
| Æon Flux | 62000000 | 5.4 | 703 |
| xXx: State of the Union | 60000000 | 4.7 | 549 |
| xXx | 70000000 | 5.8 | 1424 |
| eXistenZ | 15000000 | 6.7 | 475 |
| [REC]² | 5600000 | 6.4 | 489 |
预算 平均投票率 投票次数 标题
这对于我们来说是一个没有头绪的空数据框。让我们反转字母并再次运行它。
| title | budget | vote_average | vote_count |
|---|---|---|---|
| B-Girl | 0 | 5.5 | 7 |
| Ayurveda: Art of Being | 300000 | 5.5 | 3 |
| Away We Go | 17000000 | 6.7 | 189 |
| Awake | 86000000 | 6.3 | 395 |
| Avengers: Age of Ultron | 280000000 | 7.3 | 6767 |
| … | … | … | … |
| About Last Night | 12500000 | 6.0 | 210 |
| Aberdeen | 0 | 7.0 | 6 |
| Abduction | 35000000 | 5.6 | 961 |
| Abandoned | 0 | 5.8 | 27 |
| Abandon | 25000000 | 4.6 | 45 |
228 行 × 3 列
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