Python 我想在我的熊猫数据框中创建一列 value_counts
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I want to create a column of value_counts in my pandas dataframe
提问by user2592989
I am more familiar with R but I wanted to see if there was a way to do this in pandas. I want to create a count of unique values from one of my dataframe columns and then add a new column with those counts to my original data frame. I've tried a couple different things. I created a pandas series and then calculated counts with the value_counts method. I tried to merge these values back to my original dataframe, but I the keys that I want to merge on are in the Index(ix/loc). Any suggestions or solutions would be appreciated
我更熟悉 R，但我想看看是否有办法在 Pandas 中做到这一点。我想从我的数据框列之一创建唯一值的计数，然后将包含这些计数的新列添加到我的原始数据框中。我尝试了几种不同的方法。我创建了一个熊猫系列，然后使用 value_counts 方法计算了计数。我试图将这些值合并回我的原始数据帧，但我想合并的键在索引（ix/loc）中。任何建议或解决方案将不胜感激
Color Value
Red 100
Red 150
Blue 50
and I wanted to return something like
我想返回类似的东西
Color Value Counts
Red 100 2
Red 150 2
Blue 50 1
回答by unutbu
df['Counts'] = df.groupby(['Color'])['Value'].transform('count')
For example,
例如，
In [102]: df = pd.DataFrame({'Color': 'Red Red Blue'.split(), 'Value': [100, 150, 50]})
In [103]: df
Out[103]:
Color Value
0 Red 100
1 Red 150
2 Blue 50
In [104]: df['Counts'] = df.groupby(['Color'])['Value'].transform('count')
In [105]: df
Out[105]:
Color Value Counts
0 Red 100 2
1 Red 150 2
2 Blue 50 1
Note that transform('count')ignores NaNs. If you want to count NaNs, use transform(len).
请注意，transform('count')忽略 NaN。如果要计算 NaN，请使用transform(len).
To the anonymous editor: If you are getting an error while using transform('count')it may be due to your version of Pandas being too old. The above works with pandas version 0.15 or newer.
致匿名编辑： 如果您在使用transform('count')时遇到错误，可能是您的 Pandas 版本太旧。以上适用于 0.15 或更高版本的熊猫。
回答by Steven C. Howell
My initial thought would be to use list comprehension as shown below but, as was pointed out in the comment, this is slower than the groupbyand transformmethod. I will leave this answer to demonstrate WHAT NOT TO DO:
我最初的想法是使用如下所示的列表理解，但正如评论中指出的那样，这比groupbyandtransform方法慢。我将留下这个答案来证明什么不该做：
In [94]: df = pd.DataFrame({'Color': 'Red Red Blue'.split(), 'Value': [100, 150, 50]})
In [95]: df['Counts'] = [sum(df['Color'] == df['Color'][i]) for i in xrange(len(df))]
In [96]: df
Out[100]:
Color Value Counts
0 Red 100 2
1 Red 150 2
2 Blue 50 1
[3 rows x 3 columns]
@unutbu's method gets complicated for DataFrames with several columns which make this simpler to code. If you are working with a small data frame, this is faster (see below), but otherwise, you should use NOTuse this.
@unutbu 的方法对于具有多列的 DataFrames 变得复杂，这使得编码更简单。如果您使用的是小数据框，这会更快（见下文），否则，您应该使用NOT使用它。
In [97]: %timeit df = pd.DataFrame({'Color': 'Red Red Blue'.split(), 'Value': [100, 150, 50]}); df['Counts'] = df.groupby(['Color']).transform('count')
100 loops, best of 3: 2.87 ms per loop
In [98]: %timeit df = pd.DataFrame({'Color': 'Red Red Blue'.split(), 'Value': [100, 150, 50]}); df['Counts'] = [sum(df['Color'] == df['Color'][i]) for i in xrange(len(df))]
1000 loops, best of 3: 1.03 ms per loop
回答by 1''
df['Counts'] = df.Color.groupby(df.Color).transform('count')
df['Counts'] = df.Color.groupby(df.Color).transform('count')
You can do this with any series: group it by itself and call transform('count'):
您可以对任何系列执行此操作：将其单独分组并调用transform('count')：
>>> series = pd.Series(['Red', 'Red', 'Blue'])
>>> series.groupby(series).transform('count')
0 2
1 2
2 1
dtype: int64
回答by ZakS
One other option:
另一种选择：
z = df['Color'].value_counts
z1 = z.to_dict() #converts to dictionary
df['Count_Column'] = df['Color'].map(z1)
This option will give you a column with repeated values of the counts, corresponding to the frequency of each value in the 'Color' column.
此选项将为您提供一个包含重复计数值的列，对应于“颜色”列中每个值的频率。
