python中如果知道一个列表中有n个元素，怎么编程来创建n个新的空列表 python怎么定义一个空列表
python\u600e\u6837\u521b\u5efa\u5177\u6709\u4e00\u5b9a\u957f\u5ea6\u548c\u521d\u59cb\u503c\u7684\u5217\u88681\u3001\u9996\u5148\uff0c\u6211\u4eec\u9700\u8981\u6253\u5f00Python\u7684shell\u5de5\u5177\uff0c\u5728shell\u5f53\u4e2d\u65b0\u5efa\u4e00\u4e2a\u5bf9\u8c61member\uff0c\u5bf9member\u8fdb\u884c\u8d4b\u503c\u3002
2\u3001\u8fd9\u91cc\u6211\u4eec\u6240\u521b\u5efa\u7684\u5217\u8868\u5f53\u4e2d\u7684\u5143\u7d20\u5747\u5c5e\u4e8e\u5b57\u7b26\u4e32\u7c7b\u578b\uff0c\u540c\u65f6\u6211\u4eec\u4e5f\u53ef\u4ee5\u5728\u5217\u8868\u5f53\u4e2d\u521b\u5efa\u6570\u5b57\u4ee5\u53ca\u6df7\u5408\u7c7b\u578b\u7684\u5143\u7d20\u3002
3\u3001\u5148\u6765\u4f7f\u7528append\u51fd\u6570\u5bf9\u5df2\u7ecf\u521b\u5efa\u7684\u5217\u8868\u6dfb\u52a0\u5143\u7d20\uff0c\u5177\u4f53\u5982\u4e0b\u56fe\u6240\u793a\uff0c\u4f1a\u81ea\u52a8\u5728\u5217\u8868\u7684\u6700\u540e\u7684\u4f4d\u7f6e\u6dfb\u52a0\u4e00\u4e2a\u5143\u7d20\u3002
4\u3001\u518d\u6765\u4f7f\u7528extend\u5bf9\u6765\u6dfb\u52a0\u5217\u8868\u5143\u7d20\uff0c\u5982\u679c\u662f\u6dfb\u52a0\u591a\u4e2a\u5143\u7d20\uff0c\u9700\u8981\u4f7f\u7528\u5217\u8868\u7684\u5f62\u5f0f\u3002
5\u3001\u4f7f\u7528insert\u51fd\u6570\u6dfb\u52a0\u5217\u8868\u5143\u7d20\uff0cinsert\u4e2d\u6709\u4e24\u4e2a\u53c2\u6570\uff0c\u7b2c\u4e00\u4e2a\u53c2\u6570\u5373\u4e3a\u63d2\u5165\u7684\u4f4d\u7f6e\uff0c\u7b2c\u4e8c\u4e2a\u53c2\u6570\u5373\u4e3a\u63d2\u5165\u7684\u5143\u7d20\u3002
L = []
\u8fd9\u65f6L\u5c31\u662f\u4e00\u4e2a\u7a7a\u5217\u8868\u3002
\u9700\u8981\u6ce8\u610f\u7684\u662f\uff0c\u7a7a\u5217\u8868\u4e0d\u662fNone\uff0c\u56e0\u6b64
L = []If L is not None: # \u8fd9\u91cc\u7684\u4ee3\u7801\u603b\u662f\u4f1a\u88ab\u6267\u884c\u68c0\u67e5\u5217\u8868\u662f\u5426\u4e3a\u7a7a\u8981\u4f7f\u7528len()\uff1a
L = []if len(L): # \u8fd9\u91cc\u7684\u4ee3\u7801\u4e0d\u4f1a\u6267\u884c
def x(n):
lis= []
for i in range(n):lis.append([])
return lis
#如n=5
print(x(5))
你可以用deque()来指定列表的最大长度
list = [[] for _ in range(n)]
绛旓細def x(n): lis= [] for i in range(n):lis.append([]) return lis#濡俷=5print(x(5))
绛旓細绗竴绉嶆槸鏂板缓涓涓dict锛岄敭鏄鍒楄〃涓殑鍏冪礌锛屽兼槸缁熻鐨勪釜鏁帮紝鐒跺悗閬嶅巻list銆俰tems = ["cc","cc","ct","ct","ac"]count = {}for item in items: count[item] = count.get(item, 0) + 1print(count)#{'ac': 1, 'ct': 2, 'cc': 2}涔嬩腑鐢ㄥ埌浜嗕竴涓皬鎶宸э紝褰dict涓笉杩樻病...
绛旓細鏈変袱涓柟娉曪細浣跨敤鍒楄〃鐨刢ount鏂规硶锛屽洜涓篶ount鏂规硶鍙互鍋氬璞″唴鐨勫厓绱犺繘琛屽嚭鐜版鏁板彔鍔犺绠 list1=[1,2,3,4,4,5,6,7]list1.count(1) #缁撴灉涓1list1.count(4) #缁撴灉涓2for i in list1: if list1.count(i)>=2: print('璇ュ厓绱%s閲嶅'%i)#缁撴灉锛氳鍏冪礌4閲嶅璇ュ厓绱4閲嶅 #...
绛旓細涓句釜渚嬪瓙鍚э紝姣斿涓涓垪琛numbers=["cc","cc","ct","ct","ac"]銆備粬鐨勫厓绱犵绫绘湁3涓紙"cc","ct","ac"锛,姣忎釜绉嶇被鐨勪釜鏁板垎鍒负2锛2锛1 鐢python缁熻鍙互缂栧啓濡備笅浠ｇ爜锛況es = {} for i in numbers:res[i] = res.get(i, 0) + 1 print([k for k in res.keys()])print([v ...
绛旓細num_list=[1,2,3,4,6,7,8,4,5,6,6,6] if len(num_list)!=len(set(num_list)): print('鏈夐噸澶!!!')else: print('娌℃湁閲嶅!' )
绛旓細Python鍒楄〃鐢ㄦ硶澶у叏 鍒楄〃灏辨槸鐢ㄦ潵淇濆瓨涓缁勬暟鎹殑,浣跨敤涓嫭鍙穂7灏嗘暟鎹寘璧锋潵,鏁版嵁涔嬮棿鐢ㄨ嫳鏂囬楀彿,闅斿紑銆 1銆佸垪琛ㄧ壒鐐 濡傛灉鍒楄〃鏄涓涓灏忓濮,鍒楄〃涓鐨勬暟鎹槸涓涓釜濞冨▋,閭ｅ彲浠ヨ繖鏍锋潵鎻忚堪鍒楄〃鏄竴涓钩鏄撹繎浜虹殑灏忓濮,濂规讳細鎶婁功鍖呴噷鐨勫▋濞冩暣鐞嗙殑浜曚簳鏈夋潯,姣忎釜濞冨▋閮芥湁鑷繁鐨勭紪鍙,骞朵笖缂栧彿鏄粠0寮濮嬨傚埆浜哄湪濂逛功鍖呴噷...
绛旓細鍙互鍙傝冧笅闈㈢殑璧勬枡锛岃繃for 鐒跺悗浠嶢鐨勬暟缁勫彇鍑哄埌a锛岀劧鍚庡彇鍑 import re a = [鍒楄〃]c = []for x in a:c.append(re.findall(r'\d+',x))
绛旓細鍦≒ython涓锛屽彲浠ヤ娇鐢╥n杩愮畻绗︽潵鍒ゆ柇涓涓鏁版槸鍚﹀湪鍒楄〃涓紝绀轰緥浠ｇ爜濡備笅锛氬畾涔涓涓垪琛 numbers = [1, 2, 3, 4, 5]鍒ゆ柇3鏄惁鍦ㄥ垪琛ㄤ腑 if 3 in numbers:print("3鍦ㄥ垪琛ㄤ腑")else:print("3涓嶅湪鍒楄〃涓")鍦ㄨ繖娈典唬鐮佷腑锛屾垜浠鍏堝畾涔変簡涓涓悕涓簄umbers鐨勫垪琛紝鐒跺悗浣跨敤in杩愮畻绗﹀垽鏂3鏄惁鍦ㄥ垪琛ㄤ腑...
绛旓細count=0for elem in list1: if elem in list2: count+=1print("list1 涓湁%d涓厓绱犲湪list2涓"%count)
绛旓細1銆佸缓涓涓〃b=琛╝銆2銆佺瓑瀵筧鍋氬畬鎿嶄綔鍚庯紝鍒ゆ柇b鏄惁绛変簬a锛屼絾鍙戠幇b搴旇鏄寚鍚戜簡杩欎釜鍒楄〃锛屾棤璁哄浣曢兘浼氱浉绛夛紝python鍒楄〃娌℃湁鏇存柊鏁版嵁銆
