熊猫-逗号分隔行中的每个字符串在数据帧中出现的频率

Question

I am working with DataFrame containing two columns, one of columns contains comma separated strings, second one contains integers. I want to iterate through column with strings, save each unique string from each row, assign integer value from second column to each string. In other words,

A           B
a,b,c,d     0
a,b,c,d     10
a,b,d,e     89
a,b,d,e     111

In this example:

a = 220, b = 220, c = 10, d = 220, e = 210

I am selecting interesting columns from my csv file,

revcat = DataFrame(data, columns = ['Tag', 'Revenue']) 

This gives me ndarray with unique values in 'Tag' and transform it to another DataFrame.

uniqtag = rev1.Tag.str.split(",").apply(pd.Series).stack().unique()
tag_stack = pd.DataFrame(uniqtag)

I am stuck here. How, based on this, do I iterate through original 'Tag' column, using unique strings I found and sum values from 'Revenue' column to each 'Tag'?

Answer 1

You could do with Series.str.get_dummies , Series.mul and Series.sum :

df['A'].str.get_dummies(sep=',').mul(df['B'], axis=0).sum()

a    210
b    210
c     10
d    210
e    200

Explanation

df.A.str.get_dummies(sep=',')

This yields a DataFrame that looks like this:

   a  b  c  d  e
0  1  1  1  1  0
1  1  1  1  1  0
2  1  1  0  1  1
3  1  1  0  1  1

Then using .mul with your value column would yield:

     a    b   c    d    e
0    0    0   0    0    0
1   10   10  10   10    0
2   89   89   0   89   89
3  111  111   0  111  111

Which finally, applying .sum along index axis will give you your final output:

a    210
b    210
c     10
d    210
e    200

Answer 2

Here are the steps I'd use

1. Split on "," and use expand=True to get a dataframe back where each letter is in its own column (I'm assuming right now, based on your example, that you always have the same number of splits? Is this true?)

2. "Melt" that dataframe so that instead of having multiple columns created from each row in the original df, you have a long dataframe where each row is a letter and its index in the original df.

3. Convert from the indices to the values in the B column

4. Group by the letter and sum across B .

import pandas as pd

data = [
    ("a,b,c,d", 0),
    ("a,b,c,d", 10),
    ("a,b,d,e", 89),
    ("a,b,d,e", 111),
]
df = pd.DataFrame(data, columns=["A", "B"])

#   A       B
# 0 a,b,c,d 0
# 1 a,b,c,d 10
# 2 a,b,d,e 89
# 3 a,b,d,e 111

melted = df.A.str.split(",", expand=True).reset_index().melt(id_vars="index", value_name="A")
melted["B"] = df.B.loc[melted["index"]].values
melted.groupby("A").B.sum()

# value
# a    210
# b    210
# c    10
# d    210
# e    200

Note - I think you have the sums incorrect in the question; a few of them seem to be off by 10.