Split users.info into multiple requests when necessary

If you try to request several thousand users in one request (not sure
exactly what the limit is), it just returns HTTP 500, so split it into
multiple requests with a max of 1000 users per request.

1 files changed, 41 insertions, 1 deletions

diff --git a/slack/util.py b/slack/util.py
index 647ee02..5c45d9d 100644
--- a/slack/util.py
+++ b/slack/util.py
@@ -1,11 +1,15 @@
 from __future__ import annotations
 
-from typing import Callable, Optional
+from functools import partial
+from itertools import islice
+from typing import Callable, Iterable, Iterator, List, Optional, TypeVar
 
 import weechat
 
 from slack.shared import WeechatCallbackReturnType, shared
 
+T = TypeVar("T")
+
 
 def get_callback_name(callback: Callable[..., WeechatCallbackReturnType]) -> str:
     callback_id = f"{callback.__name__}-{id(callback)}"
@@ -18,3 +22,39 @@ def with_color(color: Optional[str], string: str, reset_color: str = "reset"):
         return f"{weechat.color(color)}{string}{weechat.color(reset_color)}"
     else:
         return string
+
+
+# From https://github.com/more-itertools/more-itertools/blob/v10.1.0/more_itertools/recipes.py#L93-L106
+def take(n: int, iterable: Iterable[T]) -> List[T]:
+    """Return first *n* items of the iterable as a list.
+
+        >>> take(3, range(10))
+        [0, 1, 2]
+
+    If there are fewer than *n* items in the iterable, all of them are
+    returned.
+
+        >>> take(10, range(3))
+        [0, 1, 2]
+
+    """
+    return list(islice(iterable, n))
+
+
+# Modified from https://github.com/more-itertools/more-itertools/blob/v10.1.0/more_itertools/more.py#L149-L181
+def chunked(iterable: Iterable[T], n: int, strict: bool = False) -> Iterator[List[T]]:
+    """Break *iterable* into lists of length *n*:
+
+        >>> list(chunked([1, 2, 3, 4, 5, 6], 3))
+        [[1, 2, 3], [4, 5, 6]]
+
+    The last yielded list will have fewer than *n* elements
+    if the length of *iterable* is not divisible by *n*:
+
+        >>> list(chunked([1, 2, 3, 4, 5, 6, 7, 8], 3))
+        [[1, 2, 3], [4, 5, 6], [7, 8]]
+
+    To use a fill-in value instead, see the :func:`grouper` recipe.
+
+    """
+    return iter(partial(take, n, iter(iterable)), [])