绘制每个 pandas 的唯一值计数的键计数 [英] Plot key count per unique value count in pandas

问题描述

我有一组数据，我希望根据这些数据绘制每个唯一ID计数的键数(em = x)，并且我我试图学习如何利用pandas.

I have a set of data from which I want to plot the number of keys per unique id count (x=unique_id_count, y=key_count), and I'm trying to learn how to take advantage of pandas.

在这种情况下:

unique_ids 1 =密钥计数2

unique_ids 1 = key count 2

unique_ids 2 =密钥计数1

unique_ids 2 = key count 1

from pandas import *
key_items = ("a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "c", "c", "c")
id_data = ("X", "X", "X", "X", "X", "X", "X", "Y", "Y", "Y", "X", "X", "X")

df = DataFrame({'keys': key_items, 'ids': id_data})

我设法通过从数据框中提取数据并对其进行重组，然后重新构建一个新的数据框来将数据整理到所需的位置.在这种情况下，最好在不带熊猫的python中完成所有操作……

I've managed to mangle the data into what I want by pulling out the data from the dataframe and restructuring it, and rebuilding a new dataframe. In this case it's probably better to do it all in python without pandas...

unique_values = defaultdict(list)
for items in df.itertuples(index=False):
    key = items[1]
    v = items[0]
    unique_values[key].append(v)

unique_values_count = {}
for k, values in unique_values.iteritems():
    unique_values_count[k] = [len(set(values))]

# reformat for plotting
key_col = ("a", "b", "c")
id_col = [unique_values_count[k][0] for k in key_col]



df2 = DataFrame({"keys":key_col, "unique_id_count": id_col})
df2.groupby("unique_id_count").size().plot(kind="bar")

有没有更好的方法可以使用初始数据帧更直接地做到这一点?

Is there a better way to do this more directly using the initial dataframe?

推荐答案

s = df.groupby("keys").ids.agg(lambda x:len(x.unique()))
pd.value_counts(s).plot(kind="bar")

这篇关于绘制每个 pandas 的唯一值计数的键计数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！