Python: Add dictionary to an existing dataframe where dict.keys() match dataframe row-CodePudding

I'm trying to add a dictionary to a 26x26 dataframe with row and column both go from a to z:

My dictionary where I want to put in the dataframe is:

{'b': 74, 'c': 725, 'd': 93, 'e': 601, 'f': 134, 'g': 200, 'h': 1253, 'i': 355, 'j': 5, 'k': 2, 'l': 324, 'm': 756, 'n': 317, 'o': 88, 'p': 227, 'r': 608, 's': 192, 't': 456, 'u': 152, 'v': 142, 'w': 201, 'x': 51, 'y': 10, 'z': 53}

I want each of my dictionary keys to match the row name of my dataframe, meaning I want this dictionary to be added vertically under the column a. As you can see, the 'a' and 'q' are missing in my dictionary, and I want them to be 0 instead of being skipped. How can I possibly achieve this?

CodePudding user response：

You can use:

df.loc[list(dic), 'a'] = pd.Series(dic)

Or:

df.loc[list(dic), 'a'] = list(dic.values())

Full example:

dic = {'b': 74, 'c': 725, 'd': 93, 'e': 601, 'f': 134, 'g': 200, 'h': 1253,
       'i': 355, 'j': 5, 'k': 2, 'l': 324, 'm': 756, 'n': 317, 'o': 88,
       'p': 227, 'r': 608, 's': 192, 't': 456, 'u': 152, 'v': 142, 'w': 201,
       'x': 51, 'y': 10, 'z': 53}

from string import ascii_lowercase
idx = list(ascii_lowercase)
df = pd.DataFrame(0, index=idx, columns=idx)


df.loc[list(dic), 'a'] = pd.Series(dic)

print(df)

output:

      a  b  c  d  e  f  g  h  i  j  ...  q  r  s  t  u  v  w  x  y  z
a     0  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
b    74  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
c   725  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
d    93  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
e   601  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
f   134  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
g   200  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
h  1253  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
i   355  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
j     5  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
k     2  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
l   324  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
m   756  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
n   317  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
o    88  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
p   227  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
q     0  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
r   608  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
s   192  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
t   456  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
u   152  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
v   142  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
w   201  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
x    51  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
y    10  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0
z    53  0  0  0  0  0  0  0  0  0  ...  0  0  0  0  0  0  0  0  0  0

[26 rows x 26 columns]

CodePudding user response：

Try iterating through the alphabet using a loop and the chr() function.

dic = {'b': 74, 'c': 725, 'd': 93, 'e': 601, 'f': 134, 'g': 200, 'h': 1253, 'i': 355, 'j': 5, 'k': 2, 'l': 324, 'm': 756, 'n': 317, 'o': 88, 'p': 227, 'r': 608, 's': 192, 't': 456, 'u': 152, 'v': 142, 'w': 201, 'x': 51, 'y': 10, 'z': 53}
for i in range(97,123):
    character = chr(i)
    if character in dic:
        df['a'][character] = dic[character]
    else:
        df['a'][character] = 0

Assuming df is your pandas data frame.