基于itertools.pairwise（非nan值）和pd.concat函数：

from itertools import pairwise

df_ = (pd.concat(df.set_index('sum').apply(
    lambda x: pd.DataFrame([[*p, x.name] for p in pairwise(x.dropna())]),
    axis=1).values, ignore_index=True)
       .set_axis(['source', 'target', 'sum'], axis=1))

source target  sum
0     IN  TOTAL    1
1  TOTAL  OUT_A    2
2  OUT_A  OUT_B    2
3  TOTAL  OUT_C    3

您可以使用DataFrame.melt通过DataFrame.dropna使用自定义函数为幻灯片窗口删除缺失值，最后创建2列DataFrame并追加sum列：

#https://stackoverflow.com/a/6822773/2901002
def window(seq, n=2):
    return [tuple(seq[i: i + n]) for i in range(len(seq) - n + 1)]
   

s = (df.melt('sum', ignore_index=False)
       .dropna(subset=['value'])
       .reset_index()
       .groupby(['index','sum'])['value']
       .apply(window)
       .explode()
       )

df1 = (pd.DataFrame(s.tolist(), columns=['source', 'target'])
         .assign(sum=s.index.get_level_values('sum')))
print (df1)
  source target  sum
0     IN  TOTAL    1
1  TOTAL  OUT_A    2
2  OUT_A  OUT_B    2
3  TOTAL  OUT_C    3

列表理解中的另一个想法：

g = (df.melt('sum', ignore_index=False)
       .dropna(subset=['value'])
       .reset_index()
       .groupby(['index','sum'])['value'])

n = 2

df1 = (pd.DataFrame([(*(k[i: i + n]), j) for (i, j), k in g for i in range(len(k) - n + 1)],
                    columns=['source', 'target','sum']))
print (df1)
  source target  sum
0     IN  TOTAL    1
1  TOTAL  OUT_A    2
2  OUT_A  OUT_B    2
3  TOTAL  OUT_C    3

Apply一个函数，它获取每行中的非空值对，explode将这些对扩展到单独的行，将每对扩展到单独的列，并将结果与原始 Dataframe 中的sum列组合。

def pairs(row):
    " Return a list of the non-null consecutive pairs in a dataframe row "
    items = row.dropna().values
    return [items[i:i+2] for i in range(len(items)-1)]

cols = ["l0", "l1", "l2", "l3"]
result = (df[cols]
          .apply(pairs, axis=1)    
          .explode()               # series of source-target pairs
          .apply(pd.Series)        # expand each pair to two columns
          .rename(columns={0: "source", 1: "target"}) # name the resulting columns
          .assign(sum=df["sum"])   # add the sum column from original df
         )

stack删除NA，然后使用groupby、zip和concat的字典理解来合并连续的值：

out = (pd.concat({k: pd.DataFrame(zip(g, g.iloc[1:]), columns=['source', 'target'])
                 for k, g in df.set_index('sum').stack().groupby(level=0)},
                names=['sum'])
         .reset_index('sum')
         .sort_index(ignore_index=True, kind='stable') # optional
       )

输出：

sum source target
0    1     IN  TOTAL
1    2  TOTAL  OUT_A
2    3  TOTAL  OUT_C
3    2  OUT_A  OUT_B