Fix string formatting

doc/source/whatsnew/v2.0.2.rst

@@ -13,7 +13,8 @@ including other versions of pandas.
 
 Fixed regressions
 ~~~~~~~~~~~~~~~~~
--
+- Fixed regression when :meth:`DataFrame.to_string` creates extra space for string dtypes (:issue:`52690`)
+
 
 .. ---------------------------------------------------------------------------
 .. _whatsnew_202.bug_fixes:
@@ -33,6 +34,7 @@ Other
 ~~~~~
 - Raised a better error message when calling :func:`Series.dt.to_pydatetime` with :class:`ArrowDtype` with ``pyarrow.date32`` or ``pyarrow.date64`` type (:issue:`52812`)
 
+
 .. ---------------------------------------------------------------------------
 .. _whatsnew_202.contributors:
 

pandas/io/formats/format.py

@@ -1392,7 +1392,10 @@ def _format(x):
         fmt_values = []
         for i, v in enumerate(vals):
             if not is_float_type[i] and leading_space or self.formatter is not None:
-                fmt_values.append(f" {_format(v)}")
+                if leading_space:
+                    fmt_values.append(f" {_format(v)}")
+                else:
+                    fmt_values.append(f"{_format(v)}")
             elif is_float_type[i]:
                 fmt_values.append(float_format(v))
             else:

pandas/tests/io/formats/test_format.py

@@ -2099,6 +2099,19 @@ def test_max_rows_fitted(self, length, min_rows, max_rows, expected):
         result = formatter.max_rows_fitted
         assert result == expected
 
+    def test_no_extra_space(self):
+        # GH 52690: Check that no extra space is given
+        # Expected Output
+        col1 = "TEST"
+        col2 = "PANDAS"
+        col3 = "to_string"
+        expected = f"{col1:<6s} {col2:<7s} {col3:<10s}"
+        # Testing
+        df = DataFrame([{"col1": "TEST", "col2": "PANDAS", "col3": "to_string"}])
+        d = {"col1": "{:<6s}".format, "col2": "{:<7s}".format, "col3": "{:<10s}".format}
+        result = df.to_string(index=False, header=False, formatters=d)
+        assert result == expected
+
 
 def gen_series_formatting():
     s1 = Series(["a"] * 100)