write fast python blog https://kongs.medium.com/writing-fast-python-5c15251124d3

a.py

%%time

cs = []
for _idx, row in df.iterrows():
    c2 = row['a']**2 + row['b']**2
    cs.append(math.sqrt(c2))

# Wall time: 667 ms

b.py

def foo(x):
    a = bar1(x)
    b = bar2(a)
    c = bar3(b)
    return c

c.py

%load_ext line_profiler
%lprun -f foo foo(x)

pt1.py

def pt1(df):
    cs = []
    for a, b in df.itertuples(index=False):
        c2 = a**2 + b**2
        cs.append(math.sqrt(c2))
    return cs

%timeit _ = pt1(df)
# 10.8 ms ± 173 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

r1.txt

Timer unit: 1e-07 s

Total time: 1.42588 s
File: <ipython-input-2-33cd5359c288>
Function: foo at line 1

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
     1                                            def foo(x):
     2         1    5139999.0 5139999.0     36.0      a = bar1(x)
     3         1    7074258.0 7074258.0     49.6      b = bar2(a)
     4         1    2044485.0 2044485.0     14.3      c = bar3(b)
     5         1         26.0     26.0      0.0       return c

r2.txt

Timer unit: 1e-07 s

Total time: 1.9012 s
File: <ipython-input-36-6fbec21419d8>
Function: pt0 at line 1

Line #      Hits         Time  Per Hit   % Time  Line Contents
==============================================================
     1                                           def pt0(df):
     2         1         21.0     21.0      0.0      cs = []
     3     10001   12287695.0   1228.6     64.6      for _idx, row in df.iterrows():
     4     10000    6535856.0    653.6     34.4          c2 = row['a']**2 + row['b']**2
     5     10000     188400.0     18.8      1.0          cs.append(math.sqrt(c2))
     6         1          5.0      5.0      0.0      return cs

v.py

%timeit _ = np.sqrt(df['a'] ** 2 + df['b'] ** 2)
# 481 µs ± 9.35 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

## if it's still too slow and we can use the numpy array as a result:
dfv = df.values
%timeit _ = np.sqrt(dfv[:, 0]**2 + dfv[:, 1]**2)
# 45.7 µs ± 179 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)