Euclidean Distance Matrix Using Pandas

the matrix can be directly created with cdist in scipy.spatial.distance:

from scipy.spatial.distance import cdist
df_array = df[["LATITUDE", "LONGITUDE"]].to_numpy()
dist_mat = cdist(df_array, df_array)
pd.DataFrame(dist_mat, columns = df["CITY"], index = df["CITY"])

for i in df["CITY"]:
    for j in df["CITY"]:
        row = df[df["CITY"] == j][["LATITUDE", "LONGITUDE"]]
        latitude = row["LATITUDE"].tolist()[0]
        longitude = row["LONGITUDE"].tolist()[0]
        df.loc[df['CITY'] == i, j] = ((df["LATITUDE"] - latitude)**2 + (df["LONGITUDE"] - longitude)**2)**0.5

df = df.drop(["CITY", "LATITUDE", "LONGITUDE"], axis=1)

This works

You can use pdist and squareform methods from scipy.spatial.distance:

In [12]: df
Out[12]:
  CITY   LATITUDE   LONGITUDE
0    A  40.745392  -73.978364
1    B  42.562786 -114.460503
2    C  37.227928  -77.401924
3    D  41.245708  -75.881241
4    E  41.308273  -72.927887

In [13]: from scipy.spatial.distance import squareform, pdist

In [14]: pd.DataFrame(squareform(pdist(df.iloc[:, 1:])), columns=df.CITY.unique(), index=df.CITY.unique())
Out[14]:
           A          B          C          D          E
A   0.000000  40.522913   4.908494   1.967551   1.191779
B  40.522913   0.000000  37.440606  38.601738  41.551558
C   4.908494  37.440606   0.000000   4.295932   6.055264
D   1.967551  38.601738   4.295932   0.000000   2.954017
E   1.191779  41.551558   6.055264   2.954017   0.000000