检查非索引列是否在 Pandas 中排序
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Check whether non-index column sorted in Pandas
提问by nick_eu
Is there a way to test whether a dataframe is sorted by a given column that's not an index (i.e. is there an equivalent to is_monotonic() for non-index columns) without calling a sort all over again, and without converting a column into an index?
有没有一种方法可以测试数据帧是否按不是索引的给定列排序（即对于非索引列，是否有等效于 is_monotonic() 的方法）而无需再次调用排序，并且无需将列转换为指数？
回答by DSM
There are a handful of functions in pd.algoswhich might be of use. They're all undocumented implementation details, so they might change from release to release:
有一些函数pd.algos可能有用。它们都是未记录的实现细节，所以它们可能会随着版本的不同而变化：
>>> pd.algos.is[TAB]
pd.algos.is_lexsorted pd.algos.is_monotonic_float64 pd.algos.is_monotonic_object
pd.algos.is_monotonic_bool pd.algos.is_monotonic_int32
pd.algos.is_monotonic_float32 pd.algos.is_monotonic_int64
The is_monotonic_*functions take an array of the specified dtype and a "timelike" boolean that should be Falsefor most use cases. (Pandas sets it to Truefor a case involving times represented as integers.) The return value is a tuple whose first element represents whether the array is monotonically non-decreasing, and whose second element represents whether the array is monotonically non-increasing. Other tuple elements are version-dependent:
这些is_monotonic_*函数采用指定 dtype 的数组和False适用于大多数用例的“timelike”布尔值。（True对于涉及以整数表示的时间的情况，Pandas 将其设置为。）返回值是一个元组，其第一个元素表示数组是否单调非递减，其第二个元素表示数组是否单调非递增。其他元组元素与版本相关：
>>> df = pd.DataFrame({"A": [1,2,2], "B": [2,3,1]})
>>> pd.algos.is_monotonic_int64(df.A.values, False)[0]
True
>>> pd.algos.is_monotonic_int64(df.B.values, False)[0]
False
All these functions assume a specific input dtype, even is_lexsorted, which assumes the input is a list of int64arrays. Pass it the wrong dtype, and it gets really confused:
所有这些函数都假定一个特定的输入数据类型 even is_lexsorted，它假定输入是一个int64数组列表。将错误的 dtype 传递给它，它会变得非常混乱：
In [32]: pandas.algos.is_lexsorted([np.array([-2, -1], dtype=np.int64)])
Out[32]: True
In [33]: pandas.algos.is_lexsorted([np.array([-2, -1], dtype=float)])
Out[33]: False
In [34]: pandas.algos.is_lexsorted([np.array([-1, -2, 0], dtype=float)])
Out[34]: True
I'm not entirely sure why Series don't already have some kind of short-circuiting is_sorted. There might be something which makes it trickier than it seems.
我不完全确定为什么 Series 还没有某种短路is_sorted。可能有一些事情使它比看起来更棘手。
回答by shx2
You can use the numpy method:
您可以使用 numpy 方法：
import numpy as np
def is_df_sorted(df, colname):
return (np.diff(df[colname]) > 0).all()
A more direct approach (like you suggested, but you say you don't want it..) is to convert to an index and use the is_monotonicproperty:
更直接的方法（如您所建议的，但您说您不想要它......）是转换为索引并使用该is_monotonic属性：
import pandas as pd
def is_df_sorted(df, colname):
return pd.Index(df[colname]).is_monotonic
回答by Konstantin
Meanwhile, since 0.19.0, there is pandas.Series.is_monotonic_increasing, pandas.Series.is_monotonic_decreasing, and pandas.Series.is_monotonic.
同时，从 0.19.0 开始，有pandas.Series.is_monotonic_increasing、pandas.Series.is_monotonic_decreasing、 和pandas.Series.is_monotonic。
