这是我想到的答案，但这些复杂的答案让我认为这是完全错误的/:))

import random

def rand5():
    return float(random.randint(0,5))

def rand7():
    random_val = rand5()
    return float(random.randint((random_val-random_val),7))

print rand7()

只需要缩放第一个函数的输出

0) you have a number in range 1-5
1) subtract 1 to make it in range 0-4
2) multiply by (7-1)/(5-1) to make it in range 0-6
3) add 1 to increment the range: Now your result is in between 1-7

通过使用滚动总数，您可以同时

保持平均分配;而且 不需要牺牲随机序列中的任何元素。

这两个问题都是简单的rand(5)+rand(5)…类型的解决方案。下面的Python代码展示了如何实现它(其中大部分是证明发行版)。

import random
x = []
for i in range (0,7):
    x.append (0)
t = 0
tt = 0
for i in range (0,700000):
    ########################################
    #####            qq.py             #####
    r = int (random.random () * 5)
    t = (t + r) % 7
    ########################################
    #####       qq_notsogood.py        #####
    #r = 20
    #while r > 6:
        #r =     int (random.random () * 5)
        #r = r + int (random.random () * 5)
    #t = r
    ########################################
    x[t] = x[t] + 1
    tt = tt + 1
high = x[0]
low = x[0]
for i in range (0,7):
    print "%d: %7d %.5f" % (i, x[i], 100.0 * x[i] / tt)
    if x[i] < low:
        low = x[i]
    if x[i] > high:
        high = x[i]
diff = high - low
print "Variation = %d (%.5f%%)" % (diff, 100.0 * diff / tt)

这个输出显示了结果:

pax$ python qq.py
0:   99908 14.27257
1:  100029 14.28986
2:  100327 14.33243
3:  100395 14.34214
4:   99104 14.15771
5:   99829 14.26129
6:  100408 14.34400
Variation = 1304 (0.18629%)

pax$ python qq.py
0:   99547 14.22100
1:  100229 14.31843
2:  100078 14.29686
3:   99451 14.20729
4:  100284 14.32629
5:  100038 14.29114
6:  100373 14.33900
Variation = 922 (0.13171%)

pax$ python qq.py
0:  100481 14.35443
1:   99188 14.16971
2:  100284 14.32629
3:  100222 14.31743
4:   99960 14.28000
5:   99426 14.20371
6:  100439 14.34843
Variation = 1293 (0.18471%)

一个简单的rand(5)+rand(5)，忽略那些返回大于6的情况，其典型变化为18%，是上面所示方法的100倍:

pax$ python qq_notsogood.py
0:   31756 4.53657
1:   63304 9.04343
2:   95507 13.64386
3:  127825 18.26071
4:  158851 22.69300
5:  127567 18.22386
6:   95190 13.59857
Variation = 127095 (18.15643%)

pax$ python qq_notsogood.py
0:   31792 4.54171
1:   63637 9.09100
2:   95641 13.66300
3:  127627 18.23243
4:  158751 22.67871
5:  126782 18.11171
6:   95770 13.68143
Variation = 126959 (18.13700%)

pax$ python qq_notsogood.py
0:   31955 4.56500
1:   63485 9.06929
2:   94849 13.54986
3:  127737 18.24814
4:  159687 22.81243
5:  127391 18.19871
6:   94896 13.55657
Variation = 127732 (18.24743%)

并且，根据Nixuz的建议，我已经清理了脚本，所以您可以提取并使用rand7…材料:

import random

# rand5() returns 0 through 4 inclusive.

def rand5():
    return int (random.random () * 5)

# rand7() generator returns 0 through 6 inclusive (using rand5()).

def rand7():
    rand7ret = 0
    while True:
        rand7ret = (rand7ret + rand5()) % 7
        yield rand7ret

# Number of test runs.

count = 700000

# Work out distribution.

distrib = [0,0,0,0,0,0,0]
rgen =rand7()
for i in range (0,count):
    r = rgen.next()
    distrib[r] = distrib[r] + 1

# Print distributions and calculate variation.

high = distrib[0]
low = distrib[0]
for i in range (0,7):
    print "%d: %7d %.5f" % (i, distrib[i], 100.0 * distrib[i] / count)
    if distrib[i] < low:
        low = distrib[i]
    if distrib[i] > high:
        high = distrib[i]
diff = high - low
print "Variation = %d (%.5f%%)" % (diff, 100.0 * diff / count)

int randbit( void )
{
    while( 1 )
    {
        int r = rand5();
        if( r <= 4 ) return(r & 1);
    }
}

int randint( int nbits )
{
    int result = 0;
    while( nbits-- )
    {
        result = (result<<1) | randbit();
    }
    return( result );
}

int rand7( void )
{
    while( 1 )
    {
        int r = randint( 3 ) + 1;
        if( r <= 7 ) return( r );
    }
}

就是这样，均匀分布，零rand5调用。

def rand7:
    seed += 1
    if seed >= 7:
        seed = 0
    yield seed

需要事先播种。

rand7() = (rand5()+rand5()+rand5()+rand5()+rand5()+rand5()+rand5())%7+1

编辑:这并不奏效。误差约为千分之二(假设是完美的rand5)。桶得到:

value   Count  Error%
1       11158  -0.0035
2       11144  -0.0214
3       11144  -0.0214
4       11158  -0.0035
5       11172  +0.0144
6       11177  +0.0208
7       11172  +0.0144

通过转换到的和

n   Error%
10  +/- 1e-3,
12  +/- 1e-4,
14  +/- 1e-5,
16  +/- 1e-6,
...
28  +/- 3e-11

似乎每增加2就增加一个数量级

BTW:上面的误差表不是通过采样产生的，而是通过以下递归关系产生的:

P [x,n]是给定n次调用rand5，输出=x可能发生的次数。

  p[1,1] ... p[5,1] = 1
  p[6,1] ... p[7,1] = 0

  p[1,n] = p[7,n-1] + p[6,n-1] + p[5,n-1] + p[4,n-1] + p[3,n-1]
  p[2,n] = p[1,n-1] + p[7,n-1] + p[6,n-1] + p[5,n-1] + p[4,n-1]
  p[3,n] = p[2,n-1] + p[1,n-1] + p[7,n-1] + p[6,n-1] + p[5,n-1]
  p[4,n] = p[3,n-1] + p[2,n-1] + p[1,n-1] + p[7,n-1] + p[6,n-1]
  p[5,n] = p[4,n-1] + p[3,n-1] + p[2,n-1] + p[1,n-1] + p[7,n-1]
  p[6,n] = p[5,n-1] + p[4,n-1] + p[3,n-1] + p[2,n-1] + p[1,n-1]
  p[7,n] = p[6,n-1] + p[5,n-1] + p[4,n-1] + p[3,n-1] + p[2,n-1]