Python 如何获取列中出现频率最高的值的个数？
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How to get the number of the most frequent value in a column?
提问by Roman
I have a data frame and I would like to know how many times a given column has the most frequent value.
我有一个数据框，我想知道给定列具有最频繁值的次数。
I try to do it in the following way:
我尝试通过以下方式做到这一点：
items_counts = df['item'].value_counts()
max_item = items_counts.max()
As a result I get:
结果我得到：
ValueError: cannot convert float NaN to integer
As far as I understand, with the first line I get series in which the values from a column are used as key and frequency of these values are used as values. So, I just need to find the largest value in the series and, because of some reason, it does not work. Does anybody know how this problem can be solved?
据我了解，第一行我得到了一系列，其中列中的值用作键，这些值的频率用作值。所以，我只需要找到系列中的最大值，但由于某种原因，它不起作用。有谁知道如何解决这个问题？
采纳答案by beardc
It looks like you may have some nulls in the column. You can drop them with df = df.dropna(subset=['item']). Then df['item'].value_counts().max()should give you the max counts, and df['item'].value_counts().idxmax()should give you the most frequent value.
看起来您的列中可能有一些空值。您可以使用df = df.dropna(subset=['item']). 然后df['item'].value_counts().max()应该给你最大计数，并且df['item'].value_counts().idxmax()应该给你最频繁的值。
回答by jonathanrocher
You may also consider using scipy's modefunction which ignores NaN. A solution using it could look like:
您也可以考虑使用mode忽略 NaN 的scipy函数。使用它的解决方案可能如下所示：
from scipy.stats import mode
from numpy import nan
df = DataFrame({"a": [1,2,2,4,2], "b": [nan, nan, nan, 3, 3]})
print mode(df)
The output would look like
输出看起来像
(array([[ 2., 3.]]), array([[ 3., 2.]]))
meaning that the most common values are 2for the first columns and 3for the second, with frequencies 3and 2respectively.
这意味着最常见的值是2第一列和3第二列3，2分别有频率和。
回答by Anton Protopopov
To continue to @jonathanrocher answer you could use modein pandas DataFrame. It'll give a most frequent values (one or two) across the rows or columns:
要继续@jonathanrocher 回答，您可以mode在 Pandas DataFrame 中使用。它将在行或列中提供最频繁的值（一或两个）：
import pandas as pd
import numpy as np
df = pd.DataFrame({"a": [1,2,2,4,2], "b": [np.nan, np.nan, np.nan, 3, 3]})
In [2]: df.mode()
Out[2]:
a b
0 2 3.0
回答by jpp
Just take the first row of your items_countsseries:
只需取items_counts系列的第一行：
top = items_counts.head(1) # or items_counts.iloc[[0]]
value, count = top.index[0], top.iat[0]
This works because pd.Series.value_countshas sort=Trueby default and so is already orderedby counts, highest count first. Extracting a value from an index by location has O(1) complexity, while pd.Series.idxmaxhas O(n) complexity where nis the number of categories.
这是有效的，因为默认情况下pd.Series.value_countshassort=True和 so已经按计数排序，最高计数在前。按位置从索引中提取值具有 O(1) 复杂度，而pd.Series.idxmax具有 O( n) 复杂度，其中n是类别数。
Specifying sort=Falseis still possible and then idxmaxis recommended:
指定sort=False仍然是可能的，然后idxmax建议：
items_counts = df['item'].value_counts(sort=False)
top = items_counts.loc[[items_counts.idxmax()]]
value, count = top.index[0], top.iat[0]
Notice in this case you don't need to call maxand idxmaxseparately, just extract the index via idxmaxand feed to the loclabel-based indexer.
请注意，在这种情况下，您不需要单独调用max和idxmax，只需通过提取索引idxmax并将其提供给loc基于标签的索引器即可。
回答by user9114146
Add this line of code to find the most frequent value
添加这行代码以查找最频繁的值
df["item"].value_counts().nlargest(n=1).values[0]
回答by Ambati Vaishnavi
The NaN values are omitted for calculating frequencies. Please check your code functionality hereBut you can use the below code for same functionality.
计算频率时省略 NaN 值。 请在此处检查您的代码功能但您可以使用以下代码实现相同的功能。
**>> Code:**
# Importing required module
from collections import Counter
# Creating a dataframe
df = pd.DataFrame({ 'A':["jan","jan","jan","mar","mar","feb","jan","dec",
"mar","jan","dec"] })
# Creating a counter object
count = Counter(df['A'])
# Calling a method of Counter object(count)
count.most_common(3)
**>> Output:**
[('jan', 5), ('mar', 3), ('dec', 2)]
