数据框 - 组合

Question

您如何结合 df_1 和 df_2 - 以实现所需的数据帧？

希望颜色方块可以快速了解所需内容。即。

为此而苦苦挣扎-感谢所有帮助/建议。谢谢。

Answer 1

您可以尝试 df_total = df1.append(df2)，但这种方式可能会将列 'race_id_legs2' 放在 'race_id_leg1' 之后，因此您需要重新组织列。

Answer 2

这是一个可能的解决方案。我确信有一个更有说服力的解决方案 - 但这很有效。

import pandas as pd
from collections import Counter
from itertools import chain

df_1=pd.DataFrame.from_dict({'name':['fred', 'fred', 'fred', 'bill', 'bill', \
'ted', 'ted', 'ted', 'ted'], 'pts':[8,4,5,7,2,3,9,8,5]})

df_2=pd.DataFrame.from_dict({'name':['pam', 'pam', 'lou', 'lou', 'lou', 'lou', \
'sam', 'sam', 'sam', 'sam'], 'pts':[5,6,5,6,5,6,5,6,5,6]})

############################################
# df_1 - setup 2 lists - first with the names (length of list for each person is 10 long)
# ....then do a list of points that is - length of list 10 long

df_1_count_of_names=list(Counter(df_1['name'].tolist()).values())

number_unique_names=df_1['name'].nunique()

count=0
start=0
end=df_1_count_of_names[count]
list_of_pts_for_each_name=[]
try:
    while count<len(df_1['pts']):

        list_of_pts_each_person=df_1['pts'][start:end].tolist()
        list_of_10_zeros=[0]*10
        pts_each_person_listof10 = list_of_pts_each_person + list_of_10_zeros[len(list_of_pts_each_person):]
        list_of_pts_for_each_name.append(pts_each_person_listof10)

        start=end
        end=start+df_1_count_of_names[count+1]
        count+=1

except IndexError:
     pass

df_1_total_list_of_pts=list(chain.from_iterable(list_of_pts_for_each_name))
# print(df_1_total_list_of_pts)

X=df_1['name'].unique().tolist()
Y=[0]*10
df_1_total_list_of_names=[]
for i in X:
    for j in Y:
        df_1_total_list_of_names.append(i)
# print(df_1_total_list_of_names)

############################################
# df_2 - setup 2 lists - first with the names (length of list for each person is 10 long)
# ....then do a list of points that is - length of list 10 long

df_2_count_of_names=list(Counter(df_2['name'].tolist()).values())

number_unique_names=df_2['name'].nunique()

count=0
start=0
end=df_2_count_of_names[count]
list_of_pts_for_each_name=[]
try:
    while count<len(df_2['pts']):

        list_of_pts_each_person=df_2['pts'][start:end].tolist()
        list_of_10_zeros=[0]*10
        pts_each_person_listof10 = list_of_pts_each_person + list_of_10_zeros[len(list_of_pts_each_person):]
        list_of_pts_for_each_name.append(pts_each_person_listof10)

        start=end
        end=start+df_2_count_of_names[count+1]
        count+=1

except IndexError:
     pass

df_2_total_list_of_pts=list(chain.from_iterable(list_of_pts_for_each_name))
# print(df_2_total_list_of_pts)

X=df_2['name'].unique().tolist()
Y=[0]*10
df_2_total_list_of_names=[]
for i in X:
    for j in Y:
        df_2_total_list_of_names.append(i)
# print(df_2_total_list_of_names)

############################################
# Now - combine the name and pts lists from df_1 and df_2 into one dataframe.

df_3=pd.DataFrame({'df_1_names':df_1_total_list_of_names, 'df_1_pts':df_1_total_list_of_pts,\
'df_2_names':df_2_total_list_of_names, 'df_2_pts':df_2_total_list_of_pts})
# print(df_3)

############################################
# Now - optional - get rid of the columns that have zeroes for both df_1 and df_2 in the
# pts columns

df_4=df_3[(df_3.df_1_pts!=0)|(df_3.df_2_pts!=0)]
print(df_4)