对列中的字符串进行排序并打印图形

您可以先sort在function中添加新列f，然后再按列对值进行排序pair of websites，最后drop_duplicates按列used_at和进行排序sort：

import pandas as pd
import itertools

df = pd.read_csv("avito_trend.csv", 
                      parse_dates=[2])


def f(df):
    dfs = []
    i = 0
    for x in [list(x) for x in itertools.combinations(df['address'].unique(), 2)]:
        i += 1
        c1 = df.loc[df['address'].isin([x[0]]), 'ID']
        c2 = df.loc[df['address'].isin([x[1]]), 'ID']
        c = pd.Series(list(set(c1).intersection(set(c2))))
        #add inverted intersection c2 vs c1
        c_invert = pd.Series(list(set(c2).intersection(set(c1))))
        dfs.append(pd.DataFrame({'common users':len(c), 'pair of websites':' and '.join(x), 'sort': i}, index=[0]))
        #swap values in x
        x[1],x[0] = x[0],x[1]
        dfs.append(pd.DataFrame({'common users':len(c_invert), 'pair of websites':' and '.join(x), 'sort': i}, index=[0]))
    return pd.concat(dfs)

common_users = df.groupby([df['used_at'].dt.year]).apply(f).reset_index(drop=True, level=1).reset_index()

common_users = common_users.sort_values('pair of websites')
common_users = common_users.drop_duplicates(subset=['used_at','sort']) 
#print common_users

graph_by_common_users = common_users.pivot(index='pair of websites', columns='used_at', values='common users')
#print graph_by_common_users

#change order of columns
graph_by_common_users = graph_by_common_users[[2015,2014]]
graph_by_common_users = graph_by_common_users.sort_values(2014, ascending=False)

ax = graph_by_common_users.plot(kind='barh', width=0.5, figsize=(10,20))
[label.set_rotation(25) for label in ax.get_xticklabels()]

rects = ax.patches 
labels = [int(round(graph_by_common_users.loc[i, y])) for y in graph_by_common_users.columns.tolist() for i in graph_by_common_users.index] 
for rect, label in zip(rects, labels): 
    height = rect.get_height() 
    ax.text(rect.get_width() + 20, rect.get_y() - 0.25 + rect.get_height(), label, fontsize=8) 

#sorting values of legend
handles, labels = ax.get_legend_handles_labels()
# sort both labels and handles by labels
labels, handles = zip(*sorted(zip(labels, handles), key=lambda t: t[0]))
ax.legend(handles, labels)     

我的图：

编辑：

评论是：

为什么要创建c_invert并x 1，x [0] = x [0]，x 1

由于年份2014和的组合2015不同4，因此第一4列和第二列中的值缺失：

used_at                                2015    2014
pair of websites                                   
avito.ru and drom.ru                 1491.0  1716.0
avito.ru and auto.ru                 1473.0  1602.0
avito.ru and e1.ru                   1153.0  1364.0
drom.ru and auto.ru                     NaN   874.0
e1.ru and drom.ru                     539.0   634.0
avito.ru and irr.ru/cars              403.0   602.0
avito.ru and am.ru                    262.0   579.0
e1.ru and auto.ru                     451.0   475.0
avito.ru and cars.mail.ru/sale        256.0   424.0
drom.ru and irr.ru/cars               277.0   423.0
auto.ru and irr.ru/cars               288.0   409.0
auto.ru and am.ru                     224.0   408.0
drom.ru and am.ru                     187.0   394.0
auto.ru and cars.mail.ru/sale         195.0   330.0
avito.ru and avtomarket.ru            205.0   299.0
drom.ru and cars.mail.ru/sale         189.0   292.0
drom.ru and avtomarket.ru             175.0   247.0
auto.ru and avtomarket.ru             162.0   243.0
e1.ru and irr.ru/cars                 148.0   235.0
e1.ru and am.ru                        99.0   224.0
am.ru and irr.ru/cars                   NaN   223.0
irr.ru/cars and cars.mail.ru/sale      94.0   197.0
am.ru and cars.mail.ru/sale             NaN   166.0
e1.ru and cars.mail.ru/sale           105.0   154.0
e1.ru and avtomarket.ru               105.0   139.0
avtomarket.ru and irr.ru/cars           NaN   139.0
avtomarket.ru and am.ru                72.0   133.0
avtomarket.ru and cars.mail.ru/sale    48.0   105.0
auto.ru and drom.ru                   799.0     NaN
cars.mail.ru/sale and am.ru            73.0     NaN
irr.ru/cars and am.ru                 102.0     NaN
irr.ru/cars and avtomarket.ru          73.0     NaN

然后，我创建所有反向组合-问题已解决。但是为什么有NaN呢？为什么组合在2014和中不同2015？

我添加到功能f：

def f(df):
    print df['address'].unique()

    dfs = []
    i = 0
    for x in [list(x) for x in itertools.combinations((df['address'].unique()), 2)]:
...
...

和输出是（为什么第一次打印两次在warning 这里描述）：

['avito.ru' 'e1.ru' 'drom.ru' 'auto.ru' 'avtomarket.ru' 'am.ru'
 'irr.ru/cars' 'cars.mail.ru/sale']
['avito.ru' 'e1.ru' 'drom.ru' 'auto.ru' 'avtomarket.ru' 'am.ru'
 'irr.ru/cars' 'cars.mail.ru/sale']
['avito.ru' 'e1.ru' 'auto.ru' 'drom.ru' 'irr.ru/cars' 'avtomarket.ru'
 'cars.mail.ru/sale' 'am.ru']

所以列表是不同的，然后组合也是不同的->我得到一些NaN值。

解决方案是对组合列表进行排序。

def f(df):
    #print (sorted(df['address'].unique()))   
    dfs = []
    for x in [list(x) for x in itertools.combinations(sorted(df['address'].unique()), 2)]:
        c1 = df.loc[df['address'].isin([x[0]]), 'ID']
        ...
        ...

所有代码是：

import pandas as pd
import itertools

df = pd.read_csv("avito_trend.csv", 
                      parse_dates=[2])

def f(df):
    #print (sorted(df['address'].unique()))   
    dfs = []
    for x in [list(x) for x in itertools.combinations(sorted(df['address'].unique()), 2)]:
        c1 = df.loc[df['address'].isin([x[0]]), 'ID']
        c2 = df.loc[df['address'].isin([x[1]]), 'ID']
        c = pd.Series(list(set(c1).intersection(set(c2))))
        dfs.append(pd.DataFrame({'common users':len(c), 'pair of websites':' and '.join(x)}, index=[0]))
    return pd.concat(dfs)

common_users = df.groupby([df['used_at'].dt.year]).apply(f).reset_index(drop=True, level=1).reset_index()
#print common_users

graph_by_common_users = common_users.pivot(index='pair of websites', columns='used_at', values='common users')

#change order of columns
graph_by_common_users = graph_by_common_users[[2015,2014]]
graph_by_common_users = graph_by_common_users.sort_values(2014, ascending=False)
#print graph_by_common_users

ax = graph_by_common_users.plot(kind='barh', width=0.5, figsize=(10,20))
[label.set_rotation(25) for label in ax.get_xticklabels()]

rects = ax.patches 
labels = [int(round(graph_by_common_users.loc[i, y])) \
for y in graph_by_common_users.columns.tolist() \
for i in graph_by_common_users.index]

for rect, label in zip(rects, labels): 
    height = rect.get_height() 
    ax.text(rect.get_width()+20, rect.get_y() - 0.25 + rect.get_height(), label, fontsize=8)

    handles, labels = ax.get_legend_handles_labels()
    # sort both labels and handles by labels
    labels, handles = zip(*sorted(zip(labels, handles), key=lambda t: t[0]))
    ax.legend(handles, labels)   

和图：