Python 绘制熊猫中每个唯一值计数的键计数
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Plot key count per unique value count in pandas
提问by monkut
I have a set of data from which I want to plot the number of keysper unique id count(x=unique_id_count, y=key_count), and I'm trying to learn how to take advantage of pandas.
我有一组数据，我想从中绘制每个唯一 id 计数（x=unique_id_count，y=key_count）的键数，我正在尝试学习如何利用.pandas
In this case:
在这种情况下：
unique_ids 1 = key count 2
unique_ids 1 = 键数 2
unique_ids 2 = key count 1
unique_ids 2 = 键数 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})
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...
通过从数据框中提取数据并对其进行重组，并重建一个新的数据框，我设法将数据改造成我想要的数据。在这种情况下，最好在没有熊猫的情况下在 python 中完成所有工作......
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")
Is there a better way to do this more directly using the initial dataframe?
有没有更好的方法可以更直接地使用初始数据框来做到这一点？
采纳答案by HYRY
s = df.groupby("keys").ids.agg(lambda x:len(x.unique()))
pd.value_counts(s).plot(kind="bar")
回答by Aziz Alto
How about just directly use value_counts()
直接用怎么样 value_counts()
pd.value_counts(df['ids']).plot.bar()
