考虑各种列条件对独特元素进行分类和计数

Question

您好，我正在使用 python 对一些数据进行分类：

Articles                                       Filename
A New Marine Ascomycete from Brunei.    Invasive Species.csv
A new genus and four new species        Forestry.csv
A new genus and four new species        Invasive Species.csv

我想知道每个“文件名”有多少个独特的“文章”。

所以我想要的输出是这样的：

Filename                             Count_Unique
Invasive Species.csv                 1
Forestry.csv                         0

另一件事，我也想得到这个输出：

Filename1                        Filename2                         Count_Common articles
Forestry.csv                     Invasive Species.csv               1

我连接了数据集，最后统计了每个“文件名”中存在的元素。

有谁愿意帮忙吗？我试过 unique(), drop_duplicates() 等，但似乎无法获得我想要的输出。

无论如何，这是我代码的最后几行：

concatenated = pd.concat(data, ignore_index =True)
concatenatedconcatenated.groupby(['Title','Filename']).count().reset_index()
res = {col:concatenated[col].value_counts() for col in concatenated.columns}
res ['Filename']

Answer 1

没有魔法。只是一些常规操作。

(1) 统计文件中的“独特”文章

编辑：添加了 (quick-and-dirty) 代码以包含带有 zero-counts

的文件名

# prevent repetitive counting
df = df.drop_duplicates()

# articles to be removed (the ones appeared more than once)
dup_articles = df["Articles"].value_counts()
dup_articles = dup_articles[dup_articles > 1].index
# remove duplicate articles and count
mask_dup_articles = df["Articles"].isin(dup_articles)
df_unique = df[~mask_dup_articles]
df_unique["Filename"].value_counts()

# N.B. all filenames not shown here of course has 0 count.
#      I will add this part later on.

Out[68]: 
Invasive Species.csv    1
Name: Filename, dtype: int64

# unique article count with zeros
df_unique_nonzero_count = df_unique["Filename"].value_counts().to_frame().reset_index()
df_unique_nonzero_count.columns = ["Filename", "count"]

df_all_filenames = pd.DataFrame(
    data={"Filename": df["Filename"].unique()}
)
# join: all filenames with counted filenames
df_unique_count = df_all_filenames.merge(df_unique_nonzero_count, on="Filename", how="outer")
# postprocess
df_unique_count.fillna(0, inplace=True)
df_unique_count["count"] = df_unique_count["count"].astype(int)
# print
df_unique_count

Out[119]: 
               Filename  count
0  Invasive Species.csv      1
1          Forestry.csv      0

(2) 统计文件间的共同文章

# pick out records containing duplicate articles
df_dup = df[mask_dup_articles]
# merge on articles and then discard self- and duplicate pairs
df_merge = df_dup.merge(df_dup, on=["Articles"], suffixes=("1", "2"))
df_merge = df_merge[df_merge["Filename1"] > df_merge["Filename2"]] # alphabetical ordering
# count
df_ans2 = df_merge.groupby(["Filename1", "Filename2"]).count()
df_ans2.reset_index(inplace=True)  # optional
df_ans2

Out[70]: 
              Filename1     Filename2  Articles
0  Invasive Species.csv  Forestry.csv         1

考虑各种列条件对独特元素进行分类和计数

classifying and counting unique elements considering various column conditions

python

counter

group-by

classification

unique