为什么我的浮点数变成一个值的 Numpy Timeseries？

Question

我在 Windows 上的 Python 2.7 中运行以下代码。

当我运行代码时（在帖子底部），我收到一个错误：

Traceback (most recent call last):

  File "<ipython-input-27-f7c0cd1d93c7>", line 75, in <module>
    if r_val_path == 0:

  File "C:\Anaconda\lib\site-packages\pandas\core\generic.py", line 698, in __nonzero__
    .format(self.__class__.__name__))

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

运行代码后，我意识到r_val、r_val_edge 和r_val_path 都是numpy timeseries'，长度等于1 或0，而不是浮点数。我什至在代码的开头明确指定了 r_val = 0

谁能告诉我为什么会在我的代码中出现这种情况？

import pandas as pd
import networkx as nx
from itertools import permutations
import numpy as np

'''import dataframes '''
df = pd.DataFrame({'fld1': ['a',    'a',    'b',    'c',    'c',    'g',    'd',    'd',    'e',    'e',    'f']
                ,  'fld2': ['b',    'c',    'f',    'd',    'g',    'd',    'e',    'b',    'c',    'f',    'b']
                , 'r_val': [0.1,    0.9,    1,  0.5,    0.5,    1,  0.8,    0.2,    0.2,    0.8,    1]})

##df of all relationships to build
flds = pd.Series(df.fld1.unique())
flds = pd.Series(flds.append(pd.Series(df.fld2.unique())).unique())

combos = []
for L in range(0, len(flds)+1):
  for subset in permutations(flds, L):
      if len(subset) == 2:
          combos.append(subset)
      if len(subset) > 2:
          break

rel_df = pd.DataFrame.from_records(data = combos, columns = ['fld1','fld2'])
rel_df['relationship'] = 0

'''build graph '''

w_edges= map(list, df.values)

DG=nx.DiGraph()
DG.add_weighted_edges_from(w_edges)

''' iterator '''
#iterate through each row of the rel_df
for index, row in rel_df.iterrows():
    #pull source and target    
    fld1_val = rel_df.fld1[index]
    fld2_val = rel_df.fld2[index]
    #pull original r_val, vlookup both fields in df         
    try:    
        r_val = df.loc[(df['fld1'] == fld1_val) & 
                       (df['fld2'] == fld2_val)]['r_val']
    except:
        r_val = 0
    #iterate through each path
    for path in nx.all_simple_paths(DG, source= fld1_val, target= fld2_val):      
        path_holder = path
        r_val_path = 0
        r_val_path = r_val_path
        #iterate through each edge in each path
        for e in np.arange(0,len(path_holder)):   
            r_val_edge = 0             
            #grab nodes in pairs
            if (e < len(path_holder)) & (
                    (path_holder[e - 1 ] <> fld1_val) &
                    (path_holder[e] <> fld2_val)):
                #grab pair of nodes
                node1 = path_holder[e - 1]   
                node2 = path_holder[e] 
                #find r_val_edge from table
                r_val_edge = df.loc[(df['fld1'] == node1) & 
                       (df['fld2'] == node2)]['r_val']
                r_val_edge = r_val_edge       
                #add r_val_edge to r_val_path
                if r_val_path == 0:
                   r_val_path = r_val_edge
                else:
                   r_val_path = r_val_path * r_val_edge 
            else:
                #path is done or path is direct connection
                # move onto the next path
                pass

        r_val += r_val_path     
        #if the r_val for the path is less than threshold then quit                
        if r_val < .00000001:
            pass

    #if the r_val for the path is less than threshold then quit                
    if r_val < .00000001:
        r_val = 0        
        pass

    #add r_val to rel_df
    rel_df.loc[(rel_df.fld1 == fld1_val) & (
                rel_df.fld2 == fld2_val),'relationship'] = r_val

Answer 1

pandas.loc 的输出通常可以返回多个位置，请参阅this link with examples