What's the difference between finding the average Euclidean distance and using inertia_ in KMeans in sklearn?
You will notice that the inertia is the sum of squared distance between each point and its nearest cluster center. Thus by removing the normalization term / data.shape[0] and instead replacing that by a squared **2 these two expressions will be equivalent

distortions_2.append(sum(np.min(cdist(data
                              , kmeanModel.cluster_centers_
                              , 'euclidean')
                              , axis = 1)) **2)


and

distortions_2.append(kmeanModel.inertia_)




For example using this artificial data source, where we generate 5 Gaussian distributions with 300 points in each.

params = [[[ 0,1],  [ 0,1]], 
          [[ 5,1],  [ 5,1]], 
          [[-2,5],  [ 2,5]],
          [[ 2,1],  [ 2,1]],
          [[-5,1],  [-5,1]]]

n = 300
dims = len(params[0])

data = []
y = []
for ix, i in enumerate(params):
    inst = np.random.randn(n, dims)
    for dim in range(dims):
        inst[:,dim] = params[ix][dim][0]+params[ix][dim][1]*inst[:,dim]
        label = ix + np.zeros(n)

    if len(data) == 0: data = inst
    else: data = np.append( data, inst, axis= 0)
    if len(y) == 0: y = label
    else: y = np.append(y, label)

num_clusters = len(params)




We will then apply KMeans using different number of clusters

k = [1,2,3,4,5,6,7,8,9,10]

inertias = []
dists = []

for i in k:
    kmeans = KMeans(i)
    kmeans.fit(data)
    inertias.append(kmeans.inertia_)
    dists.append(sum(np.min(spatial.distance.cdist(data, kmeans.cluster_centers_, 'euclidean'), axis=1)**2))

plt.plot(range(1, len(inertias)+1), inertias, label = 'Inertia')
plt.plot(range(1, len(dists)+1), dists, label = 'Distance')
plt.legend()
plt.show()