熊猫 - 以groupby的频率获取价值

Question

Can someone help me with the (possible) groupby in pandas.

Here is the df:

    easy_donor  v_fam   count
0   donor_1_NS  IGHV1   5202376
1   donor_1_NS  IGHV2   1955547
2   donor_1_NS  IGHV3   70426272
3   donor_1_NS  IGHV4   452367
4   donor_1_NS  IGHV5   4842145
5   donor_1_NS  IGHV6   490142
6   donor_1_NS  IGHV7   19708
24  donor_2_NS  IGHV1   31258603
25  donor_2_NS  IGHV2   5295899
26  donor_2_NS  IGHV3   47286417
27  donor_2_NS  IGHV4   44553802

Then I want each count as a frequency of the sum of the counts grouped by donor.

Like:

df.groupby('easy_donor').sum()['count']
easy_donor
donor_1_NS     83394639
donor_2_NS    129191591
donor_3_HS    220549762
donor_3_NS    104821016
donor_4_HS    200444923
donor_4_NS    121287306

Then each count in the original data frame divided by the groupby sum if they match the easy_donor column. Do I have to join on original dataframe?

Answer 1

尝试：

df.groupby('easy_donor')["count"].apply(lambda x: x / x.sum())

Answer 2

FORGET THIS ANSWER!!! THIS IS JUST AN IDEA. NOT VIABLE

Note that using pandas apply is unbearably slow. Instead, try using the native broadcasting .

df.groupby(by='easy_donor')['count'] * 1. / df.groupby(by='easy_donor').sum()