鍵值對的矩陣/字典

Question

我有幾個對象type dict與不同的鍵。我想創建一個表with all keys和foreach對象一行。如果一個鍵不可用，它應該是空的。

例如：

x1=dict({"a":2, "b":3}) 
x2=dict({"a":2, "b":3, "c":2}) 

，我想是這樣的：

"a","b","c" 
2,3, 
2,3,2 

Answer 1

如果使用熊貓，你可以這樣做：

import pandas as pd 
df = pd.DataFrame({k: [v] for k, v in x1.iteritems()}) 
df2 = pd.DataFrame({k: [v] for k, v in x2.iteritems()}) 

df = pd.concat((df, df2), ignore_index=True) 

# a b c 
# 0 2 3 NaN 
# 1 2 3 2 

注：iteritems()僅適用於Python 2.x.

Answer 2

作爲一般的做法（我假設你已經類型的字典，這裏的列表，你是細與具有「asciibetical」列順序）：

def EmitDictsAsCSV(list_of_dicts): 
    # First, accumulate the full set of dict keys 
    key_set = set() 
    for d in list_of_dicts: 
    key_set.update(d.iterkeys()) 

    # make a sorted list out of them 
    column_names = sorted(key_set) 
    # print the header 
    print ",".join(['"%s"' % c for c in column_names]) 

    # For each dict, loop over the columns and build a row, 
    # use the string representation of the value, if there's 
    # one, otherwise use an empty string, 
    # finish off by printing the row data, separated by commas 
    for d in list_of_dicts: 
    row_data = [] 
    for c in column_names: 
     if c in d: 
     row_data.append(str(d[c])) 
     else: 
     row_data.append("") 
    print ",".join(row_data) 

Answer 3

這裏是另一個簡單的解決方案，它不使用pandas：

all_dics = [x1, x2] 
keys = set(key for d in all_dics for key in d) # {'a', 'b', 'c'} 
dic = {key: [None]*len(all_dics) for key in keys} # {'a': [None, None], 'b': [None, None], 'c': [None, None]} 
for j, d in enumerate(all_dics): 
    for key, val in d.iteritems(): 
     dic[key][j] = val 

print dic 
# {'a': [2, 2], 'b': [3, 3], 'c': [None, 2]} 

Answer 4

這裏是一種非常粗糙的和可能低效溶液

x1=dict({"a":2, "b":3,"d":4,"e":5}) 
x2=dict({"a":2, "b":3, "c":2}) 

z = dict(x1.items() + x2.items()) 
print(z.keys()) 

x1_vals = [] 
x2_vals = [] 
for keys in z.keys(): 
    if keys in x1.keys(): 
     x1_vals.append(x1[keys]) 
    else: 
     x1_vals.append(None) 

    if keys in x2.keys(): 
     x2_vals.append(x2[keys]) 
    else: 
     x2_vals.append(None) 

print (x1_vals) 
print (x2_vals) 

輸出

['a', 'c', 'b', 'e', 'd'] 
[2, None, 3, 5, 4] 
[2, 2, 3, None, None]